.of 






01 




PLANE AND SOLID 
ANALYTIC GEOMETRY 



BY 



WILLIAM FaVbSGOOD, Ph.D., LL.D. 

PERKINS PROFESSOR OF MATHEMATICS 
IN HARVARD UNIVERSITY 



WILLIAM C^GRAUSTEIN, Ph.D. 

ASSISTANT PROFESSOR OF MATHEMATICS 
IN HARVARD UNIVERSITY 



THE MACMILLAN COMPANY 

1920 

i 

All rights reserved 



OA55 

,©7 



Copyright, 1920, 
By THE MACMILLAN COMPANY. 



Set up and electrotyped. Published November, 1920. 



g)CI.A601589 



NOV 20 1920 



NortoooH $reas 

J. S. Cushing Co. — Berwick & Smith Co. 
Norwood, Mass., U.S. A. 



X 



3 



PLANE ANALYTIC GEOMETRY 

INTRODUCTION 
DIRECTED LINE-SEGMENTS. PROJECTIONS 

Elementary Geometry, as it is studied in the high school 
to-day, had attained its present development at the time when 
Greek culture was at its height. The first systematic treat- 
ment of the subject which has come down to us was written 
by Euclid about 300 b.c. 

Algebra, on the other hand, was unknown to the Greeks. 
Its beginnings are found among the Hindus, to whom the so- 
called Arabic system of numerals may also be due. It came 
into Western Europe late, and not till the close of the middle 
ages was it carried to the point which is marked by any school 
book of to-day that treats this subject. 

When scholars had once possessed themselves of these two 
subjects — Geometry and Algebra — the next step was quickly 
taken. The renowned philosopher and mathematician, Kene 
Descartes, in his Geometrie of 1637, showed how the methods 
of algebra could be applied to the study of geometry. He 
thus became the founder of Analytic Geometry. # 

The " originals " and the locus problems of Elementary 
Geometry depend for their solution almost wholly on ingenu- 
ity. There are no general methods whereby one can be sure 
of solving a new problem of this class. Analytic Geometry, 

*Also called Cartesian Geometry, from the Latinized form of his name, 
Cartesius. 

1 



2 ANALYTIC GEOMETRY 

on the other hand, furnishes universal methods for the treat- 
ment of such problems ; moreover, these methods make pos- 
sible the study of further problems not thought of by the 
ancients, but lying at the heart of modern mathematics and 
mathematical physics. Indeed, these two great subjects owe 
their very existence to the new geometry and the Calculus. 

The question of how to make use in geometry of the nega- 
tive, as well as the positive, numbers is among the first which 
must be answered in applying algebra to geometry. The solu- 
tion of this problem will become clear in the following 
paragraphs. 

1. Directed Line-Segments. Let an indefinite straight line, 

L, be given, and let two points, A and B, be marked on L. 

. 7? r Then the portion of L which is 

1 1 1 — bounded by A and B is what is 

^ C J5 called in Plane Geometry a line- 

1 1 ~~ — • segment, and is written as AB. 

C A B ^et a tn ^ r( ^ P om t, C, be marked 

' ' ' — on L. Then three cases arise, 

"Ftp 1 

as indicated in the figure. Cor- 
responding to these three cases we have : 

(a) AB + BC = AC; 

(b) AB-CB = AC; 

(c) CB-AB=CA. 

Three other cases will arise if the original points A and B are 
taken in the opposite order on the line. Let the student 
write down the three corresponding equations. 

A unification of all these cases can be effected by means of 
an extension of the concept of a line-segment. We no longer 
consider the line-segments AB and B A as identical, but we 
distinguish between them by giving each a direction or sense. 
Thus, AB shall be directed from A to B and BA shall be 
directed from B to A, i.e. oppositely to AB. These directed 



INTRODUCTION 3 

line-segments we denote by AB and BA, to distinguish them 
from the ordinary, or undirected, line-segments. 

We may, for the moment, interpret the directed line-seg- 
ment AB as the act of walking from A to B ; then BA repre- 
sents the act of walking from B to A. With this in mind, let 
us return to Fig. 1 and consider the directed line-segments 
AB, BG, and AC. We have, in all three cases represented by 
Fig. 1, and also in the other three : 

AB + BC=AC, 

since walking from A to B and then walking from B to G is 
equivalent, with reference to the point reached, to walking 
from A to C. 

Accordingly, we unify all six cases by defining, as the sum 
of the directed line-segments AB and BG, the directed liue- 
segment AG: 

(1) AB + BC=AG. 

From this definition it follows that, if A, B, G, and D are 
any four points of L, 

(2) AB + BC+CD = AD. 

For, by (1), the sum of the first two terms in (2) is AG, and, 
by the definition, the sum of AG and CD is AD. 

Similarly, if the points M, M u M 2 , • • •, -M~ n _i, N are any points 
of L, we have 



(3) MM 1 + M,M 2 + • • • + M n . 2 M n _ x + M^N = MN. 

Given two directed line-segments on the same line or on 
two parallel lines, we say that these two directed line-segments 
are equal, if they have equal lengths and the same direction or 
sense. 

2. Algebraic Representation of Directed Line-Segments. On 

the line L let one of the two opposite directions or senses be 
chosen arbitrarily and defined as the positive direction or sense 
of L ; and let the other be called the negative direction or sense. 



4 ANALYTIC GEOMETRY 

A directed line-segment AB, which lies on L, is then called 
positive, if its sense is the same as the positive sense of L, and 
negative, if its sense is the same as the negative sense of L. 

To such a directed line-segment AB we assign a number, 
which we shall also represent by AB, as follows. If I is the 
length of the ordinary line-segment AB, then 

AB = I, if AB is a positive line-segment ; 
AB = — l, if AB is a negative line-segment. 

If AB = I, then BA = - Z ; and if JjB = - Z, then ~BA= I. 
In either case 

(1) AB + BA = or ^5 = -^Z. 

Since the act of walking from A to B is nullified by the act 
of walking from B to A, we might have arrived at equations 
(1) from consideration of the line-segments themselves, instead 
of by use of the numbers which represent them. 

It is easy to verify the fact that equations (1), (2), and (3) 
of the preceding paragraph, which relate to directed line-seg- 
ments, hold for the corresponding numbers. Consequently, 
no error or confusion arises from using the same notation AB 
for both the directed line-segment and the number correspond- 
ing to it. We shall, however, adopt a still simpler notation, 
dropping the dash altogether and writing henceforth AB to 
denote, not merely the directed line-segment or the number 
corresponding to it, but also the line-segment itself, stating 
explicitly what is meant, unless the meaning is clear from the 
context. 

Absolute Value. It is often convenient to be able to express 
merely the length of a directed line-segment, AB. The nota- 
tion for this length is | AB | ; read : " the absolute value of 
AB." 

The numerical, or absolute, value of a number, a, is 
denoted in the same way : \a\. Thus, | — 3 1 = 3. Of course, 
13|=3. 



INTRODUCTION 



3. Projection of a Broken Line. By the projection of a point 
P on a line L is meant the foot, M, of the perpendicular 
dropped from P on L. If P lies on L, it is its own projection 
on L. 

Let PQ be any directed line-segment, and let L be an arbi- 
trary line. Let M and N be respectively the projections of 
P and Q on X. The projection 
of the directed line-segment PQ 
on i shall be defined as the di- 
rected line-segment JOT", or the 
number which represents MN al- 
gebraically. Since MN— — NM, 
it follows that 

Proj. PQ = 



M 



Fig. 



N 
o 



Proj. QP. 



If PQ lies on a line perpendicular to L, the points M and 
37" coincide, and we say that the projection MN oi PQ on L is 
zero. Such a directed line-segment MN, whose end-points are 
identical, we may call a nil-segment ; to it corresponds the 
number zero. It is evident that in taking the sum of a num- 
ber of directed line-segments, any of them which are nil- 
segments may be disregarded, just as, in taking the sum of a 

set of numbers, any of them 
which are zero may be disre- 
garded. 

Consider an arbitrary 
broken line PP X P^ ■ ■ ■ P„_iQ. 
By its projection on L is 
meant the sum of the pro- 
jections of the directed line- 




M M 2 M x M, 



Fig. 6 
segments PP 1} PiP 2 , 



Pn-iQ, or 



MM X + M,M 2 + • + M n _ x N. 



This sum has the same value as MN, the projection on L 
of the directed line-segment PQ ; cf . § 1, (3) : 

MM X -f M Y M 2 + • • • + M n _ x N = MN. 



6 ANALYTIC GEOMETRY 

Hence the theorem : 

Theorem 1. Hie sum of the projections on L of the segments 
PP X , P1P2, • • -, P n -\Q of a broken line joining P with Q is equal 
to the projection on L of the directed line-segment PQ. 

If, secondly, the same points P and Q be joined by another 
broken line, PP[P' 2 • • • P' m -iQ) the projection of the latter on 
L will also be equal to MN: 

MM[ + M[M f 2 + ... + Jf^_ 1 iV= MN. 

Hence the theorem : 

Theorem 2. Given two broken lines having the same extremi- 
ties, 

PP1P2 P n -iQ ^d PP[P[ • PL-iQ- 

Let L be an arbitrary straight line. Then the sum of the pro- 
jections on L of the segments PP 1} P\Pz, • • •, P n -\Q, of ivhich the 
first broken line is made up, is equal to the corresponding sum 
for the second broken line. 



CHAPTER I 



COORDINATES. CURVES AND EQUATIONS 



1. Definition of Rectangular Coordinates. Let a plane be 
given, in which it is desired to consider points and curves. 
Through a point in this plane take two indefinite straight 
lines at right angles to each other, and choose on each line a 
positive sense. 

Let P be any point of the plane. Consider the directed 
line-segment OP. Let its projections on the two directed lines 
through be OM and ON. The numbers which represent 
algebraically these projections, 
that is, the lengths of OM and 
ON taken with the proper signs 
(cf. Introduction, § 2), are called 
the coordinates of P. We shall 
denote them by x and y : 

x = OM, y = ON, 

and write them in parentheses : 

(x, y). The first number, x, is 

known as the x-coordinate, or 

abscissa, of P; the second, y, as the y-coordinate, or ordinate, 

of P. 

The point is called the origin of coordinates. The directed 
lines through are called the axes of coordinates or the coordi- 
nate axes ; the one, the axis of x ; the other, the axis of y. It 
is customary to take the coordinate axes as in Fig. 1, the axis 
of x being positive from left to right, and the axis of y, posi- 
tive from below upward. But, of course, the opposite sense 
on one or both axes may be taken as positive, and an oblique 

7 



N 


y 


P 









a 


i 



Fig. 1 



8 ANALYTIC GEOMETRY 

position of the axes which conforms to the definition is legiti- 
mate, the essential thing being solely that the axes be taken 
perpendicular to each other. 

Every point, P, in the plane has definite coordinates, (x, y). 
Conversely, to any pair of numbers, x and ?/, corresponds a 
point P whose coordinates are (x, y). This point can be con- 
structed by layiug off OM=x on the axis of x, erecting a per- 
pendicular at M to that axis, and then laying off MP = y. 
We might equally well have begun by laying off ON= y on 
the axis of y (cf. Fig. 1), and then erected, a perpendicular to 
y that axis at N and laid off on it 

P NP = x. It shall be understood that 

the positive sense on any line parallel 
to one of the coordinate axes, such as 

■jjjL -x the perpendicular to the axis of x at 

M, shall be the same as the positive 
sense of that axis. For other lines 
of the plane there is no general principle governing the choice 
of the positive sense. 

The coordinates of the origin are (0, 0). Every point on 
the axis of x has as its ordinate, and these are the only 
points of the plane for which this is true. Hence the axis of 
x is represented by the equation 

y = 0, (axis of x). 

Similarly, the axis of y is represented by the equation 
x = 0, (axis of y). 

The axes divide the plane into four regions, called quadrants. 
The Jirst quadrant is the region included between the positive 
axis of x and the positive axis of y ; the second quadrant, the 
region between the positive axis of y and the negative axis of 
x ; etc. It is clear that the coordinates of a point in the first 
quadrant are both positive ; that a point of the second quad- 
rant has its abscissa negative and its ordinate positive ; etc. 

The system of coordinates just described is known as a sys- 
tem of rectangular or Cartesian coordinates. 



COORDINATES. CURVES AND EQUATIONS 



EXERCISES 

The student should provide himself with some squared 
paper for working these and many of the later exercises in 
this book. Paper ruled to centimeters and subdivided to mil- 
limeters is preferable. 

1. Plot the following points, taking 1 cm. as the unit : 
(a) (0,1); (6) (1,0); (c) (1,1); 
(d) (1,-1); 00 (-1,-1)5 (/) (2,-3); 

W (0,-2}); (ft) (-3.7,0); (i) (-1^, -If); 

(J) (-4,3.2); (ft) (3.24, -0.87); (J) (-1,1). 

2. Determine the coordinates of the point P in Fig. 1 when 
1 in. is taken as the unit of length ; also when 1 cm. is the 
unit of length. 

3. The same for the point marked by the period in 
"Fig. 1." 



^^"^ 2 '(x 2) y 2 ) 



2. Projections of a Directed Line-Segment on the Axes. Let 

Pit with the coordinates (x 1} yi), and P 2 : (x 2 , y 2 )* be any two 
points of the plane. Con- 
sider the directed line-seg- Pi : {x v yJ 
ment PiP 2 « It is required 
to find its projections on the 
axes. 

To do this, draw the 
broken line P 1 OP 2 . By In- 
troduction, § 3, Th. 1, the ~0 
projections of this broken Fig. 3 

line on the axes are the 

same as those of the directed line-segment PiP 2 . 
taking first the projections on the axis of x, we have : 

Proj. P^ 2 = Proj. PiO + Proj. OP 2 



Hence, 



* We shall frequently use this shorter notation 
breviation for " P 2 , with the coordinates (sc 2 , 2/2)- ' 



= -Proj. OP x + Proj. OP 2 . 

Pi '• (»2, 1/2), as an ab- 



10 ANALYTIC GEOMETRY 

But the terms in the last expression are by definition — x v 
and flj 2 - So 

(1) Proj. P1P2 on #-axis = x 2 — x 1 . 
Similarly, 

(2) Proj . P X P 2 on y-axis = y 2 - y v 

The projections of P X P 2 on two lines drawn parallel to the 
axes are obviously given by the same expressions. 

EXERCISES 

1. Plot P X P 2 when P x is the point (a) of Ex. 1, § 1, and P 2 

is (b). Determine the projections from the foregoing formulas, 
and verify directly from the figure. 

2. The same, when 

i) P 1 is (e) and P 2 is (/) ; 
ii) P x is (c) and P 2 is (d) ; 
iii) P x is (i) and P 2 is (I). 

3. Distance between Two Points. Let the points be P lf 
with the coordinates (x ly y{) 9 and P 2 : (# 2 , y 2 ). Through P, 
draw a line parallel to the axis of x and through P 2 , a line 

parallel to the axis of y\ let Q 
r^vVi) denote the point of intersection of 
these lines. Then, by the Pytha- 
gorean Theorem, 

P i :{x v y i ] 1_« (1) P^^P^+QPJ, 






Fig. 4 or 

(2) D^ = (x 2 ~x l y+(y 2 -y 1 y, 

where D denotes the distance between P x and P 2 . Hence 



(3) D^^{x 2 -x l f+{y 2 -y l )\ 

In the foregoing analysis, we have used P X Q (and similarly, 
QP 2 ) in two senses, namely, i) as the length of the ordinary 
line-segment P X Q of Elementary Geometry ; ii) as the algebraic 



COORDINATES. CURVES AND EQUATIONS 11 

expression x 2 — x x for the projection P X Q of the directed line- 
segment P\P 2 on a parallel to the axis of x. Since, however, 
these two numbers differ at most in sign, their squares are 
equal, and hence equation (2) is equivalent to equation (1). 

In particular, P X P 2 ma y be parallel to an axis, e. g. the axis 
of x. Here, y 2 = 3/1, and (3) becomes 



D = -s/{x 2 -x l f. 
The student must not, however, hastily infer that 

D = x 2 — x x . 
It may be that x 2 — x x is negative, and then * 

D = — (x 2 — #1). 

A single formula which covers both cases can be written in 
terms of the absolute value (cf. Introduction, § 2) as follows : 

(4) D=\x 2 -x x \. 

EXERCISES 



1. Find the distances between the following pairs of points, 
expressing the result correct to three significant figures. Draw 
a figure each time, showing the points and the line connecting 
them, and verify the result by actual measurement. 

(a) (2, 1) and (- 2, - 2). (b) (- 7, 6) and (2, - 3). 

(c) (13, 5) and (- 2, 5). (d) (7, 3) and (12, 3). 

(e) (4, 8) and (4, - 8). (/) (- 1, 2) and (- 1, 6). 

2. Find the lengths of the sides of the triangle whose ver- 
tices are the points (— 2, 3), (—2, — 1), (4, — 1). 

3. How far are the vertices of the triangle in question 2 
from the origin ? 

* There is no contradiction here, or conflict with the ordinary laws of 
algebra. For, the V- S ig n always calls for the positive square root, — that 
being the definition of the symbol, — and we must see to it in any given 
case that we fulfill the contract. 



12 



ANALYTIC GEOMETRY 



4. Find the lengths of the diagonals of the convex quadri- 
lateral whose vertices are the points (4, 1), (1,3), (—3,1), 
(-2,-1). 

4. Slope of a Line. By the slope, X, of a line is meant the 
trigonometric tangent of the angle, 0, which the line makes 

with the positive axis of x : 

(1) A. = tan $. 

To find the slope of the line, 
let Pj, with the coordinates (x lt y x ), 
and P 2 : (x 2 , y 2 ) he the extremities 
of any directed line-segment PiP 2 
on the line. Then 




^0 



Fig. 5 



(2) 
or 
(3) 



tan0 = 



QPo 


_2/2" 


-2/i 


PiQ 


x 2 - 


-a?! 


__2/2" 


-2/i. 





x 2 — a?! 



If, instead of P x P 2y we had taken its opposite, P 2 Pi, we 
should have obtained for X the value (y L — ^/(^i — #2)- But 
this is equal to the value of X given by (3). Thus, X is the 
same, whether the line is directed in the one sense or in the 
opposite sense. Hence we think of X as the slope of the line 
without regard to sense. 

Variation of the /Slope. Consider the slopes, A, of different 
lines, L, through a given point, P. When L is parallel to the 
axis of x, X has the value zero. When L rotates as shown in 
the figure, X becomes positive and increases steadily in value. 
As L approaches the vertical line L\ X becomes very large, 
increasing without limit. 

When L passes beyond L\ X changes sign, being still nu- 
merically large. As L continues to rotate, X increases alge- 
braically through negative values. Finally, when L has again 
become parallel to the axis of x, X has increased algebraically 
through all negative values and becomes again zero. 



COORDINATES. CURVES AND EQUATIONS 13 



When L is in the position of L', is 90° and tan = A. is 
undefined, that is, has no valne. Hence V has no slope. One 
often sees the expression : tan 90° = oo, and, in accordance with 
it, one might write here, A = oo . This does not mean that V has 
a slope, which is infinite, for " infinity " is not a number. It 
is merely a brief and 
symbolic way of describ- 
ing the behavior of X for 
a line L, near to, but not 
coincident with V ; it 
says that for such a line 
A. is numerically very 
large : and further that, 
when the line L ap- 
proaches V as its limit, 
A increases numerically 
without limit, — that is, FlG 

increases numerically be- 
yond any preassigned number, as 10,000,000 or 10,000,000 !, and 
stays numerically above it. 




The Angle 6. In measuring the angle from one line to an- 
other, it is essential, first of all, to agree on which direction 
of rotation shall be considered as positive. We shall take 
always as the positive direction of rotation that from the posi- 
tive axis of x to the positive axis of y ; so that the angle from 
the positive axis of x to the positive axis of y is -f 90°, and 
not - 90°. 

The complete definition of is, then, as follows : TJie slope- 
angle 6 of a line is the angle from the positive axis of x to the 
direction of the line. There are in general two positive values 
for less than 360° ; if the smaller of them is denoted by 6, the 
other is 180° -f- 9. Which of these angles is chosen is imma- 
terial, since tan (180° -f- 0) = tan • this result is in agreement 
with the previous one, to the effect that the slope pertains to 
the undirected line without regard to a sense on it. 



14 ANALYTIC GEOMETRi' 

The student should now draw a variety of lines, indicating 
for each the angle 0, and assure himself that the deduction of 
formula (3) holds, not merely when the quantities x 2 — x l and 
V-2 — V\ are positive, but also when one or both are negative. 

Right-Handed and Left-Handed Coordinate Systems. For the 
choice of axes in Fig. 1, the positive direction for angles is 
the counter-clockwise direction. But for 

x such a choice as is indicated in the 

present figure, — a choice equally legiti- 
mate, — it is the clockwise sense which is 
positive. 

The above formulas apply to either 

Fig. 7 system of axes. The first system is 

called a right-handed system ; the other, 

a left-handed system. We shall ordinarily use a right-handed 

system. 

Problem. To draiv a line through a given point having a 
given slope. In practice, this problem is usually to be solved 
on squared paper. The solution will be. sufficiently clearly 
indicated by an example or two. 

Example 1. To draw a line through the point (—2, 3) hav- 
ing the slope — 4. 

Proceed along the parallel to the #-axis through the given 
point by any convenient distance, as 1 unit, toward the left.* 
Then go up the line through this point, parallel to the y-axis, 
by 4 times the former distance, — here, 4 units. Thus, a sec- 
ond point on the desired line is determined, and the line can 
now be drawn with a ruler. 

If the given point lay near the edge of the paper, so that 
the above construction is inconvenient, it will do just as well 
to proceed from the first point toward the right by 1 unit, and 
then down by four units. 

* The student will follow these constructions step by step on a piece 
of squared paper. 



COORDINATES. CURVES AND EQUATIONS 15 

Example 2. To draw a line through the point (1.32, 2.78) 
having the slope .6541. 

Here, it is clear that we cannot draw accurately enough to 
be able to use the last significant figure of the given slope. 
Open the compasses to span 10 cm. (if the squared paper is 
ruled to cm.) and lay off a distance of 10 cm. to the right on a 
parallel to the a>axis through the given point. This parallel 
need not actually be drawn. Its intersection, Q, with the cir- 
cular arc is all that counts, and this point, Q, can be estimated 
and marked. Its distance above the axis of x will be 1 cm. 
and 3.2 mm. The error of drawing will be of the order of the 
last significant figure, namely, more than ^ mm. and less than 
.5 mm. 

Next, open the compasses to span 6 cm. and 5.4 mm. Put 
the point of the compasses on Q, and lay off the above dis- 
tance, 6.54 cm., on a parallel through Q to the y-axis and 
above Q. The point B, thus found, will be a second point on 
the desired line, which now can be drawn. 

EXERCISES 

1. The points P i} P 2 , P 3 , with the coordinates (2, 5), (7, 3), 

( — 3, 7) respectively, lie on a line. Show that the value for 
the slope of the line as given by equation (3) is the same, no 
matter which two of the three points are used in obtaining it. 

2. Find the slopes of the sides of the triangle of Ex. 2, § 3. 

3. Find the angles which the sides of that triangle make 
with the axes, and hence determine the angles of the triangle. 

4. Show that the points (- 2, - 3), (5, - 4), (4, 1), (- 3, 2) 
are the vertices of a parallelogram. 

5. Draw a line through the point (1, —2) having the 
slope 3. 

6. Draw a line through the point (—2, — 1) having the 
slope — 1J. 

7. Draw a line through the point (—1.32, 0.14) having the 
slope - .2688. 



16 



ANALYTIC GEOMETRY 



ffJ-'CV^)! 



P:(oc,y) 



5. Mid-Point of a Line-Segment. Let P lt with the coordi- 
nates (a,*!, 2/1), and P 2 : (x 2 , y 2 ) be the extremities of a line-seg- 
ment. It is desired to find the 
jjj ; ^2^2) coordinates of the point P which 
bisects PiP 2 . 

Let the coordinates of P be 
(#, y). It is evident that the 
directed line-segment P X P is 
—x equal to the directed line-seg- 
ment PP 2 . Hence the projec- 
tion of P}P on the axis of x, or 
Xi, mnst eqnal the projection of PP 2 on that axis, or x 2 — x : 



Fig. 8 



Hence 



Xi + x 2 

2 



Similar considerations apply to the projections on the axis 
of y, and consequently 

v= yi + y* . 
2 

We have thus obtained the following result : The coordinates 
(x } y) of the point P which bisects the line-segment PiP 2 are given 
by the equations : 



(1) 



x x + x 2 
2 ] 



= Vi + 

2 



EXERCISES 

1. Determine the coordinates of the mid-point of each of 
the line-segments given by the pairs of points in Ex. 1, § 3. 
Draw figures and check your answers. 

2. Find the mid-points of the sides of the triangle mentioned 
in Ex. 2, § 3, and check by a figure. 

3. Determine the coordinates of the mid-point of the line 
joining the points (a + b, a) and (a — 6, b). 

4. Show that the diagonals of the parallelogram of Ex. 4, 
§ 4 bisect each other. 



COORDINATES. CURVES AND EQUATIONS 17 

6. Division of a Line-Segment in Any Ratio.* Let it be re- 
quired to find the coordinates (x, y) of the point P which 
divides the line-segment P X P 2 in an arbitrary ratio, rrii/m 2 : t 

P 1 P == m l 

PP 2 m 2 ' 

Obviously the projections of P X P and PP 2 on the axis of x 
must be in the same ratio, mjm 2 , and hence 

x — x x _ m t 
x 2 — x m 2 

On solving this equation for x, it is found that 

„ _ ■m 2 x 1 + miX 2 
ra 2 + m x 

Similar considerations, applied to the projections on the 
axis of y, lead to the corresponding formula for y, and thus 
the coordinates of P are shown to be the following : 

(1) x = ^2-£i + m&z m 2 y l ^-m l y 2 

m 2 -f- m 1 m 2 + ^i 

If m x and m 2 are equal, these formulas reduce to those of 
§5. 

External Division. It is also possible to find a point P on 
the indefinite straight line through P 1 and P 2 and lying outside 
the line-segment P^P}, which makes 

P 1 P = m 1 

P 2 P m 2 

where mx and m 2 are any two unequal positive numbers. Here, 
a?! — x _ m 1 
x 2 — x m 2 

* This paragraph may well be omitted till the results are needed in 
later work. 

t The given numbers mi and m 2 may be precisely the lengths P\P 
and PP 2 ; but in general they are merely proportional respectively to 
them, i.e. they are these lengths, each multiplied by the same positive or 
negative number. 



18 ANALYTIC GEOMETRY 

On solving this equation for x and the corresponding one for 
?/. we find, as the coordinates of the point P, the following : 

(2) x = m 2- v i - m i^2 y = m&i — r "h! h . 

m 2 — mi Wh — ^1 

The point P is here said to divide the line P^P 2 externally in 
the ratio 7n l /m 2 ; and, in distinction, the division in the earlier 
case is called internal division. Both formulas, (1) and (2), 
can be written in the form (1) if one cares to consider external 
division as represented by a negative ratio, m l /m 2 , where, then, 
one of the numbers m 1? ra 2 is positive, the other, negative. 

EXERCISES 

1. Find the coordinates of the point on the line-segment 
joining (— 1, 2) with (5, — 4) which is twice as far from the 
first point as from the second. Draw the figure accurately 
and verify. 

2. Find the point on the line through the points given in 
the preceding problem, which is outside of the line-segment 
bounded by them and is twice as far from the first point as 
from the second. 

3. Find the point which divides internally the line-segment 
bounded by the points (3, 8) and (— 6, 2) in the ratio 1 : 5, and 
lies nearer the first of these points. 

4. The same question for external division. 

7. Curve Plotting. Equation of a Curve. Since the subject 
of graphs is now very generally taught in the school course 
in Algebra, most students will already have met some of the 
topics taken up on the foregoing pages, and moreover they 
will have plotted numerous simple curves on squared paper 
from given equations. Thus, in particular, they will be famil- 
iar with the fact that all the points whose coordinates satisfy 
a linear equation, i.e. an equation of the first degree, like 

(1) 2a>-3y-l = 0, 



COORDINATES. CURVES AND EQUATIONS 19 



lie on a straight line, though they may never have seen a 
formal proof. 

A number of points, whose coordinates satisfy equation (1), 
can be determined by giving to x simple values, computing 
the corresponding values of y from (1), and then plotting the 
points (x, y). Thus 

if x = 0, y — — i, and the point is (0, — £) , 
if x = 1, y = -§-, and the point is (1, -J) ; 
if x = 2, y = 1, and the point is (2, 1) ; 
if it* = — 1, y = — 1, and the point is (— 1, — 1); 
etc. 

Of course, if it is known that (1) represents a straight line, 
— i.e. that all the points whose 
coordinates satisfy (1) lie on a 
straight line, — it is sufficient 
to determine two points as above, 
and then to draw the line 
through them. 

This process of determining a 
large number of points whose 
coordinates satisfy a given equa- 
tion and then passing a smooth 

curve through them is known as " plotting a curve * from its 
equation." 

The mathematical curved defined by an equation in x and 
y consists of all those points and only those points whose co'ordi- 
nateSy when substituted for x and y in the equation, satisfy it. 

Suppose, for example, that the equation is 

(2) y = x\ 

The point (2, 4) lies on the curve denned by (2), because, when 

* In Analytic Geometry the term curve includes straight lines as well 
as crooked curves. 

•*• This curve is sometimes called the locus of the equation. 




Fig. 9 



20 ANALYTIC GEOMETRY 

x is set equal to 2 and y is set equal to 4 in (2), the resulting 
equation, 

4 = 4, 

is true. We say, equation (2) is satisfied by the coordinates of 
the point (2, 4), or that the point (2, 4) lies on the curve (2) 
On the other hand, the point (— 1, 2), for example, does not 
lie on the curve defined by (2). For, if we set x — — 1 and 
y = 2, equation (2) becomes 

2 = 1. 

This is not a true equation; i.e. equation (2) is not satisfied 
by the coordinates of the point (— 1, 2), and so this point does 
not lie on the curve (2). 

Equation of a Curve. A curve may be determined by simple 
geometric conditions ; as, for example, that all of its points 
be at a distance of 2 units from the origin. This is a circle 
with its center at the origin and having a radius of length 2. 

It is easy to state analytically the condition which the coor- 
dinates of any point (x, y) on the circle must satisfy. Since 
by § 3 the distance of any point (x, y) from the origin is 

Vz 2 + 2/2, 

the condition that (x, y) be a point of the curve is clearly this, 
that 

V^TF=2, 

or that 

(3) a^ + 2/ 2 = 4. 

Equation (3) is called the equation of the curve in question. 

TJie equation of a curve is an equation in x and y which is 
satisfied by the coordinates of every point of the curve, and by 
the coordinates of no other point. 

In this book we shall be engaged for the most part in find- 
ing the equations which represent the simpler and more im- 
portant curves, and in discovering and proving, from these 
equations, properties of the curves. 



COORDINATES. CURVES AND EQUATIONS 21 

Nevertheless, the student should at the outset have clearly 
in mind the fact that any equation between x and y, like 

y = a?, y = logx, y=sinx, 

represents a perfectly definite mathematical curve, which he 
can plot on paper. Moreover, he is in a position to determine 
whether, in the case of a chosen one of these curves, a given 
point lies on it. He will find it desirable to plot afresh a few 
simple curves, and to test his understanding of other matters 
taken up in this paragraph by answering the questions in the 
following exercises. 

EXERCISES 

1. What does each of the following equations represent? 
Draw a graph in each case. 

(a) x = 2; (c) x-y = 0; (e) 2 a? - 3y + 6 = 0; 

(6) 2y + 3 = ; (d) 2x + 5y = ; (/) 5x + Sy - 4 = 0. 

Plot the following curves on squared paper. 

2. y = x\ 

Take 2 cm. or 1 in. as the unit of length. Use a table of 
squares. 

3. y 2 = x. 

Take the same unit as in question 2 and use a table of square 
roots. 

4. Show that, when one of the curves of Exs. 2 and 3 has 
been plotted from the tables, the other can be plotted from 
the first without the tables. 

Work the corresponding exercises for the following curves. 

5. y = x 3 . 6. y = y/x. 7. y' 2 = x 2 . 8. y z = x 1 . 

9. Plot the curve 

y = \og 10 x 

from a table of logarithms for values of x from 1 to 10, taking 
1 cm. as the unit. 



22 ANALYTIC GEOMETRY 

10. Which of the straight lines of Ex. 1 go through the 
origin ? 

11. Show that the curve 

(a) y = sin x 

goes through the origin. 

Do the curves 

(b) y = tan x, (c) y = cos x, 

go through the origin ? 

12. Do the following points lie on the curve 

xy = l? 
(a) (-1,-1); (b) (-1,1); (c) (f, f) ; 

(*) (-*-*); W (h -2); (/) (0,1). 

13. Find the equations of the following curves. 

(a) The line parallel to the axis of x and 8 units above it. 

(b) The line parallel to the axis of y and 1-f units to the 
left of it. 

(c) The line bisecting the angle between the positive axis 
of y and the negative axis of x. 

(d) The circle, center in the origin, radius p. 

(e) The circle, center in the point (1, 2), radius 3. 

Arts. (x-l)*+(y-2y = 9. 

8. Points of Intersection of Two Curves. Consider, for ex- 
ample, the problem of finding the point of intersection of the 
lines 

L: 2x-3y = 4:, 

L': 3a + 4?/ = -11. 

Let (x x , 2/ x ) be the coordinates of this unknown point, P lm 
Any point P, with the coordinates (x, y), which lies on L, has 
its x and y satisfying the first of the above equations. Hence, 
in particular, since P x lies on L, x x and y x must satisfy that 
equation, or 

(1) 2a* -3* -4 



COORDINATES. CURVES AND EQUATIONS 23 

Similarly, a point P : (x, y), which lies on V, has its x and 
y satisfying the second of the above equations. Hence, in 
particular, since P, lies on V, x v and y x must satisfy that equa- 
tion, or 
(2) 3^+4^ = -11. 

Thus it appears that the two unknown quantities } x x and y u 

satisfy the two simultaneous equations, (1) and (2). Hence 
these equations are to be solved as simultaneous by the 
methods of Algebra. 

2^-3^ = 4, 4 

3;r 1 + 4?/ 1 = -ll, 3 

To do this, eliminate y x by multiplying the first equation 
through by 4, the second by 3, and then adding : 

17^ = — 17, or' x 1 = — 1. 

On substituting this value of a^ in either equation (1) or (2), 
the value of y 1 is found to be : y x — — 2. Hence P x has the 
coordinates (— 1, — 2). 

The equations (1) and (2) are the same, except for the sub- 
scripts, as the equations of the given lines, L and V. Hence 
we may say : To find the coordinates of the point of intersection 
of two lines given by their equations, solve the latter as simul- 
taneous equations in the unknown quantities, x and y, by the 
methods of Elementary Algebra. 

The generalization to the case of any two curves given by 
their equations is obvious. The equations are to be regarded 
as simultaneous equations between the unknown quantities, x and 
y, and solved as such. 

The student should observe that the letters " x " and " y " 
have totally different meanings when they appear as the co- 
ordinates of a variable point in the equation of a curve, and when 
they represent unknown quantities in a pair of simultaneous 
equations. In the first case, they are variables, and a pair of 
values, (x, y), which satisfy equation L will not, in general, 
satisfy V. In the second case, x and y are constants, the 



24 ANALYTIC GEOMETRY 

coordinates of a single point, or of several points ; but of 
isolated and not variable points. 

EXERCISES 

Determine the points of intersection of the following curves. 
Check your results by plotting the curves and reading off as 
accurately as possible the coordinates of the points of 
intersection. 

1. The straight lines (a) and (d) of Ex. 1, § 7. 

2. The straight lines (c) and (e) of Ex. 1, § 7. 

3. The straight lines (e) and (/) of Ex. 1, § 7. 

4 [ y 2 = 4», 

\x + y=3. Ans. (1,2), (9, -6). 

5 |^ + ^ = 13, 6< \x* + f- = a\ 
\xy = 6. \ x + y = 0. 

[a*+jj" = 25, 8 (y* + 6* = 0, 

} 4 a; 2 + 36 ^ = 144. \2x + y = l. 

\x* + tf = 2, 

|^ = 1. Ans. (1,1), (-1,-1). 

10. Show that the curves 

y = \og w x, x + y = l, 

intersect in the point (1, 0). 

11. Show that the curves 

# 2 + 2/ 2 = 25, 3x-±y=0, 

intersect in the point (4, 3), and also in (—4, — 3). 

EXERCISES ON CHAPTER I 

1. Show that the points (2, 0), (0, 2), (1 + V3, 1 + V3) are 
the vertices of an equilateral triangle. 

2. Prove that the triangle with vertices in the points (1, 8), 
(3, 2), (9, 4) is an isosceles right triangle. 



COORDINATES. CURVES AND EQUATIONS 25 

3. Show that the points (-1, 2), (4, 10), (2,3), and 
( — 3, — 5) are the vertices of a parallelogram. 

4. Given the points A, B, C with coordinates (—7, —2), 
(_ i^i, 0), (5, 3). By proving that 

AB + BC=AC, 

show that the three points lie on a line. 

5. Show that the three points of the previous problem lie 
on a line by proving that AB and AC have the same slope. 

6. Prove that the two points (5, 3) and (—10, — 6) lie on 
a line with the origin. 

7. Prove that the two points (x lf y{), (x 2 , y 2 ) lie on a line 
with the origin when, and only when, their coordinates are 

proportional : 

x 1 :y 1 = x 2 : y 2 . 

8. Determine the point on the axis of x which is equidis- 
tant from the two points (3, 4), (—2, 6). 

9. If (3, 2) and ( — 3, 2) are two vertices of an equilateral 
triangle which contains within it the origin, what are the co- 
ordinates of the third vertex ? 

10. If (3, —1), (—4, —3), (1, 5) are three vertices of a 
parallelogram and the fourth lies in the first quadrant, find 
the coordinates of the fourth. Arts. (8, 7). 

11. If P is the mid-point of the segment PiP 2 , and P and 
P 1 have coordinates (8, 17), (—5, — 3) respectively, what are 
the coordinates of P 2 ? 

12. If P divides the segment P X P 2 in the ratio 2 : 1, and P 1 
and P have coordinates (3, 8) and (1, 12) respectively, deter- 
mine the coordinates of P 2 . Ans. (0, 14). 

13. Find the ratio in which the point B of Ex. 4 divides 
the segment AC of that exercise. Ans. 2 : 3. 

14. A point with the abscissa 6 lies on the line joining the 
two points (2, 5), (8, 2). Find its ordinate. 



26 ANALYTIC GEOMETRY 

Suggestion. Determine the ratio in which the point divides 
the line-segment between the two given points. 

15. Prove that the sum of the squares of the distances of 
any point in the plane of a given rectangle to two opposite 
vertices equals the sum of the squares of the distances from it 
to the two other vertices. 

Suggestion. Choose the axes of coordinates skillfully. 

16. If D is the mid-point of the side BC of a triangle ABC, 
prove that 

AB* + AC' = 2 AD 2 + 2 BB\ 

17. Show that the lines joining the mid-points of opposite 
sides of a quadrilateral bisect each other. 

18. Prove that the lines joining the mid-points of adjacent 
sides of a quadrilateral form a parallelogram. 

19. Prove that, if the diagonals of a parallelogram are equal, 
the parallelogram is a rectangle. 

20. If two medians of a triangle are equal, show that the 
triangle is isosceles. 



CHAPTER II 



THE STRAIGHT LINE 



1. Equation of Line through Two Points. Let P y : (x ly y t ) 
and P 2 : (#2> ft) be two given points, and let it be required to 
find the equation of the line through 
them. 

The slope of the line, by Ch. I, 
§ -t, is 

ft — ft 



X'o 




Fig. 1 



Let P, with the coordinates (a;, #), 

be any point on the line other than P v 

the line is also given by 

y-ft. 



Then the slope of 



Hence 
(1) 



x x 



y-ft = ft-ft 

a; — a*! # 2 — #! 

Conversely, if P : (x, y) is any point whose coordinates 
satisfy equation (1), this equation then says that the slope of 
the line P X P is the same as the slope of the line P X P 2 and 
hence that P lies on the line PiP 2 . 

A more desirable form of equation (1) is obtained by multi- 
plying each side by (x — x l )/(y 2 — yi). We then have : 



(i) 



x - "-'l _ y 

X 2 



ft 



®1 2/2 - ft 

Equation (I) is satisfied by the coordinates of those points and 
only those points which lie on the line PiP 2 . Consequently. 

27 



28 ANALYTIC GEOMETRY 

by Ch. 1, § 7, (I) is the equation of the line through the two given 
points. 

Example 1. Find the equation of the line which passes 
through the points (1, — 2) and (— 3, 4). 
Here 

^i = I? Vi — — ^ and x 2 = — 3, y 2 = 4= 
By (I) the equation of the line is 

s-l- y-(-2) or X -l_y + 2 



_3_1 4-(-2)' -4 " 6 

On clearing of fractions and reducing, the equation becomes 

3x + 2y + l=:0. 

Let the student show that, if (x 1} y x ) had been taken as 
(—3, 4) and (x 2 , y 2 ) as (1, — 2), the same equation would have 
resulted. 

Example 2. Find the equation of the line passing through 
the origin and the point (a, b). 

Here, (x lf ^i) = (0, 0) and (x 2 , y 2 ) = (a, b), and (I) becomes 

- = % or bx~ay = 0. 
a b 

Lines Parallel to the Axes. In deducing (I) we tacitly as- 
sumed that 

y-i — Vi ¥= and x 2 — x l ^ ; 

for otherwise we could not have divided by these quan- 
tities. 

If y 2 — y l = 0, the line is parallel to the axis of x. Its 
equation is, then, obviously 

(2) y = jfcJ 

Similarly, if x 2 — x l — 0, the line is parallel to the axis of y 
and has the equation 

(3) x — x v 

These two special cases are not included in the result em- 



THE STRAIGHT LIXE 29 

bodied in equation (I). We see, however, that they are so 
simple, that they can be dealt with directly.* 

Example 3. Find the equation of the line passing through 
the two points (—5, 1) and (—5, 8). 

It is clear from the figure that this line is parallel to the 
axis of y and 5 units distant from it to the 
left. Accordingly, the abscissa of every point 
on it is — 5 ; conversely, every point whose 
abscissa is — 5 lies on it. Therefore, its equa- 
tion is 



(-5,8) 



<-5,D 





x = — 5, or x + 5 = 0. fig. 2 



EXERCISES f 
Draw the following lines and find their equations. 

1. Through (1, 1) and (3, 4). Ans. Sx - 2y - 1 = 0. 

2. Through (5, 3) and (- 8, 6). 

3. Through (0, - 5) and (- 2, 0). Ans. 5x+2y + 10 = 0. 

4. Through the origin and (—1, 2). 

5. Through the origin and (— 2, — 3). 

6. Through (2, - 3) and (- 4, - 3). Ans. y + 3 = Q. 

* It is not difficult to replace (I) by an equation which holds in all 
cases, — namely, the following : 

(I') O2 - 2/1) (x - xi) = (x 2 - x{) (y - y x ) . 

We prefer, however, the original form (I) . For (I) is more compact 
and easier to remember, and the special cases not included in it are best 
handled without a formula. 

t In substituting numerical values for (Xi, y{) and (z 2 , 2/2) iu (I), the 
student will do well to begin with a framework of the form 

x- _y — 



and then fill in each place in which X\ occurs ; next, each place in which 
2/1 occurs ; and so on. When x x or y x is negative, substitute it first in 
parentheses ; thus, if x x = — 3, begin by writing 

x-(-S) = y- 
-(-3) - 



30 ANALYTIC GEOMETRY 

7. Through (0, 8) and (0, - 56). 

8. Through (5, 3) and parallel to the axis of y. 

9. Through (5, 3) and parallel to the axis of x. 

10. Through (a, b) and (6, a). Ans. x -\- y = a + b 

11. Through (a, 0) and (0, b). ^ ® +1 =1 

a 6 

2. One Point and the Slope Given. Let it be required to find 
the equation of the line which passes through a given point 
Pi : (x u 2/ x ) and has a given slope, X. 

If P : (x, y) be any second point on the line, the slope of the 
line will be, by Ch. I, § 4, 

.V-ffi ,. 
x— x x 

But the slope of the line is given as A.. Hence 



x — % 
or 

(ii) y-yi = K®- x i)- 

The student can now show, conversely, that any point, whose 
coordinates (x, y) satisfy (II), lies on the given line. Hence 
(II) is the equation of the line passing through the given point 
and having the given slope. 

Example. Find the equation of the line which goes through 
the point (2, — 3) and makes an angle of 135° with the posi- 
tive axis of x. 

Here, X = — 1 and (x lt y 1 ) = (2, — 3), and hence, by (II), the 
equation of the line is 

2, + 3 = -l(a>-2), 
or x -f y + 1 = 0. 

Slope-Intercept Form of Equation. It is frequently conven- 
ient to determine a line by its slope X, and the ^-coordinate 
of the point in which it cuts the axis of ?/. 




THE STRAIGHT LINE 31 

Here, Xi = ; and, if we denote y x by the letter b, (II) be= 
comes 
(III) y = \x + b. 

This is known as the slope-intercept form of the equation 
of a straight line ; b is known as y 

the intercept of the line on the axis 
of y- 

Example. Find the equation of the 
line which makes an angle of 60° with 
the axis of x and whose intercept on the w FlQ 3 
axis of y is — 2. 

Since \ = V3 and b = — 2, the equation is 
y = V3a;-2. 

EXERCISES 
Draw the following lines and find their equations, 

1. Through (— 4, 5) and with slope — 2. 

Ans. 2# + y + 3 = 0. 

2. Through (3, 0) and with slope f. 

3. Through (f, — +) and with slope — -§. 

4. Through the origin and making an angle of 60° with the 
axis of x. 

5. Through (— 4, 0) and making an angle of 45° with the 
axis of y. 

6. With intercept 1 on the axis of y and with slope — f. 

Ans. 3x + 2y-2 = 0. 

7. With intercept \ on the axis of y and making an angle 
of 30° with the axis of x. 

8. With slope — 1 and intercept — c on the axis of y. 

9. With slope a/b and intercept 6 on the axis of y. 

Ans. ax—by + 62=0. 

3. The General Equation of the First Degree. Let there be 
given an arbitrary line of the plane. If the line is parallel 



32 ANALYTIC GEOMETRY 

to neither axis, its equation is of the form (T), § 1, — an equa- 
tion of the first degree in x and y. If the line is parallel to 
the axis of x, its equation is of the form y = yi, — a special 
equation of the first degree in x and y, in which it happens 
that the term in x is lacking. Similarly, if the line is parallel 
to the axis of y, its equation is of the form x = x l9 — an equa- 
tion of the first degree which lacks the term in y. Conse- 
quently, we can say : The equation of every straight line is of the 
first degree in x andy. 

Given, conversely, the general equation of the first degree in 
x and y y namely 

(1) Ax + By+C = 0, 

where A, B, C are any three constants, of which A and B are 
not both zero ; * this equation represents always a straight line. 

The Case B=£0. In general, B will not be zero and we 
can divide equation (1) through by it : 

and then solve for y : An 

y B B 

But this equation is precisely of the form (III), § 2, where 

B' B 

Therefore, it represents a straight line whose slope is — A/B 

and whose intercept on the axis of y is — C/B. 

The Case B = 0. If, however, B is zero, the equation (1) 

becomes . „ _ 

Ax+C=0. 

Now, A cannot be zero, since the case that both A and B are 
zero was excluded at the outset. We can, therefore, divide by 
A and then solve for x : 

„_ A - 

* In dealing with equation (1), now and henceforth, we shall always 
assume that A and B are not both zero. 



THE STRAIGHT LINE 33 

This is the equation of a straight line parallel to the axis 
of y, if C =£ 0. If C — 0, it is the equation of this axis. 

This completes the proof that every equation of the first 
degree represents a straight line. In accordance with this 
property, such an equation is frequently called 
a linear equation. 

Example. What line is represented by the 
equation 

6x + 3y + l = 0? 

If we solve for y, we obtain 



y = -2x 



1\ 


.V 

XX 


1 \ 


<o>-i> 



Fig. 4 



Hence the equation represents the line of slope 

— 2 with intercept — \ on the axis of y. From these data we 

may draw the line. 

EXERCISES 

Find the slopes and the intercepts on the axis of y of the 
lines represented by the following equations. Draw the lines. 

1. 4z -|-2?/ -1 = 0. 4. 2x-y = l. 

2. 7«-f 8^ + 5 = 0. 5. y = 0. 

3. 2x — oy = Q. 6. x = 3 — y. 

Find the slopes of each of the following lines. 

7. —x + 2y = 7. Ans. i 11. 2y — 3 = 0. 

8. x — y + 1. 12. 2x = Sy. 

9. 3 — 2x = oy. 13. x = 5y + l. 
10. 2x— 3i/ = 4. 14. foe + ay = a&. 

4. Intercepts. In the preceding paragraph we learned to 
plot the line represented by a given equation, from the values 
of its slope and its intercept on the axis of y, as found from 
the equation. It is often simpler, however, in the case of a 
line which cuts the axes in two distinct points, to determine 
from the equation the coordinates of these two points and then 
to plot the points and draw the line through them. 



34 ANALYTIC GEOMETRY 

The point of intersection of a line, for example, 

(1) 2x- Sy + 4 = 0, 

with the axis of x has its y-coordinate equal to 0. Conse- 
quently, to find the ^--coordinate of the point, we have but to 
set y = in the equation of the line and solve for x. In this 
case we have, then, 



2a + 4 = 0, or 



tL°k- 



Similarly, the ^-coordinate of the point of intersection of 
the line with the axis of y is 0, and its ^/-coordinate is obtained 
by setting x = in the equation of the line and solving for y. 
In the present case this gives 

- Sy +4 = 0, or y = f • 

The points of intersection of the line (1) with the axes of 
coordinates are, then, (—2, 0) and (0, -J). We now plot these 
points and draw the line through them. 
fo^A) We recognize the number | as the 

intercept of the line (1) on the axis of y ; 
■*-« the number — 2 we call the intercept on 
the axis of x. We have plotted the line 
(1), then, by finding its intercepts. 
In general, the intercept of a line on the axis of x is the 
.f-coordinate of the point in which the line meets that axis. The 
intercept on the axis of y is similarly defined. These defini- 
tions admit of extension to any curve. Thus, the circle of Ch. I, 
§ 7, has two intercepts on the axis of x, namely, + 2 and — 2. 
An axis or a line parallel to an axis has no intercept on that 
axis. Every other line has definite intercepts on both axes, 
and these intercepts determine the position of the line unless 
they are both zero, that is, unless the line goes through the 
origin. 

EXERCISES 

Determine the intercepts of the following lines on each of 
the coordinate axes, so far as such intercepts exist, and draw 
the lines. 



THE STRAIGHT LINE 35 

1. 2x + 3y- 6 = 0. 6. 2x + 3 = 0. 

Ans. 3, 2. Ans. — 1£, none. 

2. a — y + 1 = 0. 7. 8 — 5y = 0. 

3. 3a -5?/ + 10 = 0. 8. .r=0. 

4. 5« 4- 7# -h 13 = 0. 9. x 4- y = a. 

5. 2a — 3?/ = 0. 10. 2«a — 3by = ab. 

5. The Intercept Form of the Equation of a Line. Given a 
line whose position is determined by its intercepts. Let the 
intercept on the axis of x be a, and let that on the axis of y be 
b. To find the equation of the line in terms of a and b. 

Since one point on this line is (a, 0) and a second is (0, 6), 
we have, by (I), § 1, 

x— a _ y — 
0-a~b-0' 
or 

(IV) -+!=*• 

Only lines which intersect the axes in two points that are 
distinct can have their equations written in this form. A line 
through the origin is an exception, because one or both its 
intercepts are zero and division by zero is impossible. Also a 
line parallel to an axis is an exception, since it has no inter- 
cept on that axis. 

EXERCISES 

Find the equations of the following lines. 

1. "With intercepts 5 and 3. 

2. "With intercepts — 2-J and 8. 

3. With intercepts f and — -f. 

4. The diagonals of a square lie along the coordinate axes, 
and their length is 2 units. Find the equations of the four 
sides (produced). 

An*, x 4- y = 1 ; aj— y = 1 ; — as 4- y = 1 ; — x — y = 1. 



36 ANALYTIC GEOMETRY 

5. A triangle has its vertices at the points (0, 1), (— 2, 0), 
(1, 0). Draw the triangle and find the equations of its sides 
(produced). Use formula (IV), when possible. 

6. A triangle has its vertices at the points (a, 0), (b, 0), 
(0, c). Find the equations of the sides (produced). 

7. A line goes through the origin and the mid-point of that 
side of the triangle of Ex. 5 which lies in the first quadrant. 
Find its equation. 

8. Find the equations of the lines through the origin and 
the respective mid-points of the sides of the triangle of Ex. 6. 

6. Parallel and Perpendicular Lines. Parallels. Given two 
lines oblique to the axis of y y so that both have slopes. The 
lines are parallel if, and only if, they have equal slopes. For, 
if they are parallel, their slope angles, and hence their slopes, 
are equal ; and conversely. 

Example 1. To find the equation of the line through the 
point (1, 2) parallel to the line 
(1) 3x-2y + 6=0. 

The slope of the line (1) is J- . The required line has the 
same slope and passes through the point (1, 2). By (II), § 2, 
its equation is 

2,-2^1(^-1), 
or 

3x-2y + 1 = 0. 

If the given line is parallel to the axis of y, it has no slope and 
hence the method of Example 1 is inapplicable. But then the 
required line must also be parallel to the axis of y and its 
equation can be written down directly. For example, if the 
given line is 3 a? -f 8 = 0, and there is required the line parallel 
to it passing through the point (—8, 2), it is clear that the 
required line is parallel to the axis of y and 8 units to the left 
of it, and consequently has the equation x — — 8, or x -f- 8 = 0. 

Perpendiculars. Given two lines oblique to the axes, so that 
both have slopes, neither of which is zero. The lines are per- 



THE STRAIGHT LINE 37 

pendicular if, and only if, their slopes, k x and a 2 , are negative 
reciprocals of one another : 

(2) a 2 = - f , or Xi = — f, Xi ^= 0, A 2 =£ 0. 

Ai A 2 

For, if the lines are perpendicular, one of their slope angles, 
#! and 2 , may be taken as 90° greater than the other, viz. : 

2 = X + 90°, 
and hence 

Ao = tan 2 = tan (6 1 + 90°)= - cot X = — = -i, 

tan 6 X Ai 

or 

Conversely, if this last equation is valid, the steps can be 
retraced and the lines shown to be perpendicular to each 
other. 

Example 2. To find the equation of the line through the 
point (1, 2) perpendicular to the line (1). 

The slope of (1) is f . Hence the required line has the slope 
— -|. We have, then, to find the equation of the line through 
the point (1, 2) with slope — f. By (II), § 2, this equation is 

2 ,_2 = -|( !B -l), 

or 

2^ + 3?/- 8 = 0. 

If the given line is parallel to an axis, it has no slope or its 
slope is zero. In either case, equation (2) and the method of 
Example 2 are inapplicable. But then the required line must 
be parallel to the other axis and it is easy to write its equation. 
Suppose, for example, that the given line is 2y — 3 = 0, — a 
line parallel to the axis of x, — and that the required line per- 
pendicular to it is to go through the point (3, 5). Then this 
line must be parallel to the axis of y and at a distance of 
3 units to the right of it. Consequently, its equation is 
x -3 = 0. 

The methods of this paragraph are applicable to all problems 



38 ANALYTIC GEOMETRY 

in which it is required to find the equation of a line which 
passes through a given point and is parallel, or perpendicular, 
to a given line. 

EXERCISES 

In each of the following exercises find the equations of the 
lines through the given point parallel and perpendicular to 
the given line. 

Line Point 

1. 4a-8y = 5, (-1, -3). 

Arts, x — 2^ — 5 = 0; 2x + y + 5 = 0. 

2. x-y = l, (0,0). 

3. 5x + 13^-3 = 0, (2, -1). 

4. 3x + 5y =0, (5,0). 

5. 2x = 3, (5, -6). 

6. V22/ + 7r = 0, (-2,0). Ans. y = 0; a; + 2 = 0. 

7. l-a; = 0, (0, tt). 

8. Find the equations of the altitudes of the triangle of 
§ 5, Ex. 5. 

9. Find the equations of the perpendicular bisectors of the 
sides of the triangle of § 5, Ex. 5. 

10. Show that the equation of the line through the point 
( x i> Vi) parallel to the line 

(3) Ax + By = C 

is Ax + By — Ax 1 + By x . 

11. Show that the equation of the line through the point 
( x i> Vi) perpendicular to the line (3) of Ex. 10 is 

Bx — Ay — Bx x — Ay v 

7. Angle between Two Lines. Let L x and L 2 be two given 
lines, whose slopes are, respectively, 

Ai = tan l9 and k 2 = tan $ 2 . 
To tind the angle, <f>, from Li to L 2 . 



THE STRAIGHT LINE 



39 



Since 

<ft = 2 — $ ly 

it follows from Trigonometry that 

. . tan 6 2 — tan 0, 

tan <f> = s i- , 

1 + tan 6i tan 2 

and hence that 



(i) 



tan <f> = 



A 2 — Aj 




Fig. 6 



That is, it is the 



1 + Ai* 2 

The angle (f> is the angle from L\ to L 2 . 
angle through which L x must be rotated in the positive sense, 
about the point A, in order that it coincide with L 2 . In par- 
ticular, we agree to take it as the smallest such angle, always 
less, then, than 180° : < <j> < 180°.* 

If L x and L 2 are perpendicular, then, by (2), § 6, A 2 = — 1/Ax 
and 1 -f- A r \ 2 = 0. Consequently, cot <£, which is equal to the 
reciprocal of the right-hand side of (1), has the value zero, and 
so $ = 90°. 

Example. Let L x and L 2 be given by the equations, 
A: 4a; — 2y -f- 7 = 0, 

L 2 : l2x + 4:y-5 = 0. 

Here X x = 2 and A 2 = — 3, and (1) becomes 

3-2 



tan<£ = 



= 1. 



1-6 
Hence the angle <j> from Li to L 2 is 45°. 

In deducing (1) it was assumed that L x and 
L 2 both have slopes. If this is not the case, 
at least one of the lines is parallel to the axis 
of y and no formula is needed. The angle 
<f> may be found directly. Suppose, for ex- 
ample, that Li and L 2 are, respectively, 

x + 2 = and x — y = 1. 

* The figure shows L\ and Z 2 as intersecting lines, but formula (1) and 
the deduction of it are valid also in case L\ and L 2 are parallel. In this 




Fig. 7 



40 ANALYTIC GEOMETRY 

Then L x is parallel to the axis of y, and L 2 is inclined at an 
angle of 45° to the positive axis of x, since A 2 = 1. Conse- 
quently, <£ = 135°. 

EXERCISES 

In each of the following exercises determine whether the 
given lines are mutually parallel or perpendicular, and in case 
they are neither, find the angle from the first line to the second. 

1. x + 2y = 3, x + 2y = 4. 

2. 2#-?/ + 5 = 0, 4a> — 2^-7 = 0. 

3. x — y = 1, x + y = 2. 

4. x + 2y + ll = 0, 6a; -3^-4 = 0. 
5c 3a — ?/ = 0, 2a; + ?/ = 0. 

6= a; + 2y + 1 = 0, 2a; + y-l = 0. 

7. 4a? + 3?/ = 3, 9x — 3y = 5. 

8. 2a; — 3y = l, a? — 3 = 0. 

9. x + y = 0, 2/ = 0. 

10. 2a?-3?/ + l = 0, Sx-4,y-l = 0. 

11. By the method of this paragraph determine each of the 
three angles of the triangle whose sides have the equations 

aj-2y-6 = 0, 2a; + y -4 = 0, 3x-y + 3 = 0. 

Check your results by addiug the angles. 

12. Prove that if L x and L 2 are represented by the equations 
Lx : A& + B L y + C x = 0, 

L 2 : ^ + A?/ + C 2 = 0, 

then tan ^> = 



^Bs - A 2 B X 



A X A 2 + JBiBa 

What can you say of L x and L 2 if A^2 — A 2 B X = ? If 
A^2 + ^A = 0? 

case, we take the angle from L\ to L 2 as 0° — , not as 180°, as is conceiv- 
able. Hence arises the sign < (less than or equal to) in the place in 
which it stands in the double inequality. 



THE STRAIGHT LINE 



41 



13. Show that the formula of Ex. 12 for tan <j> is valid even 
if one or both of the lines has no slope, i.e. is parallel to the 
axis of y. 

8. Distance of a Point from a Line. Let P:(x ly tfa) be a 
given point and let 

L : Ax + By+C = 

be a given line. To find the dis- 
tance, D, of P from L. 

Drop a perpendicular from P 
on the axis of x, and denote the 
point in which it cuts L by Q. 
The abscissa of Q is a^. Denote 
its ordinate by y q . Then 

Since Q : (x h y^) lies on L, its coordinates satisfy the equation 
of L ; thus 

Ax 1 + By Q +C=0. 

Solving this equation for y Q , we find : 

Axt_+ C 




Fig. 8 



y« 



Hence 
(1) 



QP 



B 



Ax x + By x + C 
B 



Let 6 be the slope-angle of L and form the product QPcos0. 
One or both of the factors of this product may be negative, 
according to the positions of P and L* But always the 
numerical value of the product is equal to the distance D : 

(2) Z)=|QPcos0|. 

This is clear in case P and L are situated as in Fig. 8 ; 

* There are four essentially different positions for P and X, for L may 
have a positive or a negative slope, and P may lie on the one or on the 
other side of L. 



42 



ANALYTIC GEOMETRY 



the student should draw the other typical figures and show 
that for them, also, (2) is valid. 
Since the slope of L is 

X=rtan0=^-— , 

we have 

A> + B* 



sec 2 = 1 -f- tan • 6 = 



B* 



Consequently, 

(3) cosfl = :j- — - B 

V^l 2 + B 2 

It is immaterial to us which sign in (3) is the proper one. 
For, according to (2), we have now to multiply together the 
values of QP and cos 6, as given by (1) and (3), and take the 
numerical value of the product. The result is the desired 
formula : 



n 


m \Axi + By x +C\ 




■VA' + B* 


or 


, Axi + Byi + C 



(4) 

D = ±- 

V^l 2 4- -S 2 

where, in the second formula, that sign is to be chosen which 
makes the right-hand side positive. 

Example. The distance of the point (3, — 2) from the line 
?>x + ky- 7 = 



is 



j9=r l3-3 + 4(-2)-7l = 



14. 



V3 2 + 4 2 V25 5 

The deduction of formula (4) involves division by B and 

hence tacitly assumes that B=^0, i.e. that L is not parallel to 

the axis of y. The formula holds, however, even when L is 

parallel to the axis of y. For, in this case it is clear from a 



figure that 



D=K+^ 



and (4) reduces precisely to this when B = 0. 



THE STRAIGHT LINE 



43 



Ans. 2f. 



Ans. 2V10, or 6.32. 



EXERCISES 
In each of the first seven exercises find the distance of the 
given point from the given line. 

Point Line 

1. (5,2), 3s-4?/ + 6 = 0. 

2. (2,3), 5a; + 12y 4- 2 = 0. 

3. (6, -1), 3a?-2/ + l = 0. 

4. (3,4), 3a + 5 = 0. 

5. (-2,-5), 2/ = 0. 

6. Origin, x + y — 1 = 0. 

7. Origin, Sx + 2y - 6 = 0. 

8. Find the lengths of the altitudes of the triangle with 
vertices in the points (2, 0), (3, 5), (—1, 2). 

9. Area of a Triangle. Let a triangle be given by means of 
its vertices (x 1} y x ), (x 2 , y 2 ), (a? 3 , y 3 ). 
To find its area. 

Drop a perpendicular from one 
of the vertices, as (x 3 , y 3 ), on the 
opposite side. Then the required 
area is 

where D denotes the length of 
the perpendicular and E, the length of the side in question. 
By Ch. I, § 3, we have 




<*t,y,) 



<*i^i) 



Fig. 9 



E = -V(x 2 -x l y+(y 2 -y i y. 

D is the distance of (x 3 , y 3 ) from the line joining (x 1} y{) 
and (#2, y 2 ) . The equation of this line, as given by (I) or (I'), 
§ 1, may be put into the form : 

(2/2 — 2/i)» - ( x 2 — V\)y — »i2/2 + *Wi = 0. 
Consequently, by (4), § 8, we find : 

D _ ± (2/2 ~ 2/1)3.3 - (S2 - Si>7 3 ~ 3q,y 2 + *Wl . 
V(x 2 -x 1 y+(y 2 -y 1 y 



44 ANALYTIC GEOMETRY 

Thus 

A = ±±[(ij 2 - 2/i)x 3 - (x 2 - xi)y s - x x y 2 + x 2 yQ. 

The result may be written more symmetrically in either of 
the forms 

(1) A = ± ■§-[(<*?! - x 2 )y 3 + (x 2 - a? 3 )2/i + ( x s ~ ®i)##l 
or 

(2) A = ± |[(2/! - 2/ 2 )<b 3 + (2/ 2 - 2/3)^ 4- (2/3 - 2/i)^]j 

where in each case that sign is to be chosen which makes the 
right-hand side positive. 

EXERCISES 
Find the area of the triangle whose vertices are in the 
points 

1. (1,2), (-1,2), (-2,1). 

2. (5,3), (-3,4), (-2,-1). 

3. (1, 2), (2, 1), (0, 0). 

Find the area of the triangle whose sides lie along the lines 

4. x — y = 0, x + y = 0, 2x + y-3 = 0. 

5. 2x + y~r6 = 0, x-y + 3 = 0, x-2y-8 = 0. 

6. Find the area of the convex quadrilateral whose vertices 
are in the points (4, 2), (- 1, 4), (- 3, - 2), (5, - 8). 

7. What do formulas (1) and (2) become when one of the 
vertices, say (x 3 , y z ), is in the origin ? 

Ans. A — ± ^(x x y 2 — x 2 y Y ). 

10. General Theory of Parallels and Perpendiculars. Iden- 
tical Lines.* The line through the point (a?i, y x ) parallel to 
the line 

(1) Ax + By = C, 

has the equation, according to § 6, Ex. 10, 
Ax + By = Ax x + By x . 

* The discussion in the class-room of the subjects treated in this and the 
following paragraph may well be postponed until the need for them arises. 



THE STRAIGHT LINE 45 

This equation is of the form 

(2) Ax + By=C, 

since the constant Ax ± + By x may be denoted by the single 
letter C 

Conversely, equations (1) and (2), for C =£ 0, always repre- 
sent parallel lines. For, if B=^0, the lines have the same 
slope, — A B ; if B = 0. A cannot be zero, and the lines are 
parallel to the axis of y and hence to each other. 

Theorem 1. Tico lines are paraMel when and only when their 
equations can be written in the forms (1) and (2), inhere C =£ C. 

The line through the point (x 1: y^, perpendicular to the 
line (1), has the equation (§ 6, Ex. 11) : 

Bx — Ay = Bx\ — Ay lt 

and this equation is of the form 

(3) Bx-Ay=&. 

Let the student show, conversely, that equations (1) and (3) 
always represent perpendicular lines. 

Theorem 2. Two lines are perpendicular when and only 
when their equations can be written in the forms (1) and (3). 

The equations of two parallel lines can always be written 
in the forms (1) and (2). But they need not be so written. 
Thus the lines, 

2x-y=-l, 

6x-3y = 2, 

are parallel, though the equations are not in the forms (1) and 
(2). The coefficients of the terms in x and y are not respec- 
tively equal. They are, however, proportional : 2 : 6= — 1 : — 3. 
This condition holds in all cases. For the two lines 

W- A^ + B^ + C^O, 

L 2 : A& -f- B 2 y + C 2 = 0, 

we mav state the theorem : 



46 ANALYTIC GEOMETRY 

Theorem 3. TJie lines L x and L 2 are parallel * if and only if 

A 1 :A 2 = B 1 : B 2 . 

For, L x and L 2 are parallel if and only if the angle <f> be- 
tween them, as defined in § 7, is zero ; bnt, according to § 7, 
Ex. 12, cj>, or better, tan <£, is zero, when and only when 
A X B 2 — A 2 B X = 0. But this equation is equivalent to the pro- 
portion A x : A 2 = B x : B 2 . 

As a second consequence of § 7, Ex. 12, we obtain the fol- 
lowing theorem. 

Theorem 4. The lines L x and L 2 are perpendicular if and 
only if 

Identical Lines. Two equations do not have to be identically 
the same in order to represent the same line. For example, 
the equations, 

2a-y + l = 0, 

6a;- 3^ + 3 = 0, 

represent the same line. The corresponding constants in 
them are not equal, but they are proportional. We have, 
namely, 

2:6 = -1: -3 = 1:3, 

or, what amounts to the same thing, 

2:-l:l=6: -3:3. 
This condition is general. We formulate it as a theorem : 
Theorem 5. The lilies Li and L 2 are identical if and only if 

A\ > A 2 = -Z>i : B 2 = Ci : C 2l 
or A l .B 1 :C l = A 2 .B 2 : C 2 . 

For, Ly and L 2 are the same line when and only when they 
have the same slope and the same intercept on the axis of y, 
that is, when and only when 

_^i = _4? and _5i = _^, 
Bi B 2 Bi B 2 

* Or, in a single case, identical. Of . Th. 5. 



THE STRAIGHT LINE 47 

or Ai:A 2 =Bi:B 2 and B l :B 2 =Ci'.C 2 , 

or, finally, A x : A 2 = B x : B 2 = Ci : 2 . 

This proof assumes that i^ =£ and i? 2 ^ 0. The proof, 
when this is not the case, is left to the student. 

EXERCISES 

1. Prove Th. 3 directly, without recourse to the results 
of §7. 

2. The same for Th. 4. 

See also Exs. 15, 16, 17, 18 at the end of the chapter. 

11. Second Method of Finding Parallels and Perpendiculars. 

Problem 1. To find the equation of a line parallel to the 
given line 

(1) Ax + By=C y 

and satisfying a further condition. 

By § 10, Th. 1, the desired equation can be written in the 
form _ 

(2) Ax + By=C, 

where C" is to be determined by the further condition. 

Example. Consider the first example treated in § 6. In 
this case the equation of the desired line can be written in the 
form 

3x-2y = k, 

where we have replaced the Q' of (2) by k. The "further 
condition," by means of which the value of k is to be deter- 
mined, is that the line go through the point (1, 2). Hence 
x = 1, y = 2 must satisfy the equation of the line, or 

3.1-22 = *. 
Consequently, k = — 1, and the equation of the line is 
3x-2y + l = 0. 

Problem 2. To find the equation of a line perpendicular to 
the given line (1) and satisfying a further condition. 



48 ANALYTIC GEOMETRY 

By § 10, Th. 2, the desired equation can be written in the 
form 

(3) Bx-Ay = C", 

where C is to be determined by applying the further condition. 
This condition does not always have to be that the line 
should go through a given point. It may be any single con- 
dition, not affecting the slope of the line, which it seems de- 
sirable to apply. We give an example illustrating the method 
in such a case. 

Example. To find the equation of the line perpendicular to 
2x-y-± = Q 

and cutting from the first quadrant a triangle whose area is 16. 
Equation (3) may, in this case, be written as 

(4) x + 2y = k. 

We are to determine k so that the line (4) cuts from the first 
quadrant a triangle of area 16. The intercepts of the line (4) 
are k and ^k, and hence the area of the triangle in question is 
Ik 2 . Accordingly, \k 2 = 16, and k = ± 8. But the line cuts 
the first quadrant only if k is positive, and so we must have 
7; = 8. The equation of the desired line is, then, 
x + 2y-8 = 0. 

EXERCISES 

1. Work Exs. 1-4, 8, 9 of § 6 by this method. 

2. There are two lines parallel to the line 

x — 2y = 6 

and forming with the coordinate axes triangles of area 9. 
Find their equations. 

3. Find the equations of the lines parallel to the line of 
Ex. 2 and 3 units distant from it. 

Suggestion. Write the equation of the required line in the 
form (2) and demand that the distance from it of a chosen 
point of the given line be 3. 



THE STRAIGHT LINE 49 

4. Find the equations of the lines parallel to the line 

5 X + 122/-3 = 
and 2 units distant from the origin. 

5. The same as Ex. 2, if the lines are to be perpendicular, 
instead of parallel, to the given line. 

6. The same as Ex. 4, if the lines are to be perpendicular, 
instead of parallel, to the given line. 

7. A line is parallel to the line 3x-{- 2y — 6=0, and forms 
a triangle in the first quadrant with the lines, 

x — 2 y = and 2 x — y = 0, 

whose area is 21. Find the equation of the line. 

Ans. 3a + 2?/- 28=0. 

EXERCISES ON CHAPTER II 

1. Find the equation of the line whose intercepts are twice 
those of the line 2x — 3y — 6 = 0. 

2. Find the equation of the line having the same intercept 
on the axis of x as the line V3# — y — 3 = 0, but making with 
that axis half the angle. 

3. Find the equation of the line joining the point (3, — 2) 
with that point of the line 2x — y = 8 whose ordinate is 2. 

4. A perpendicular from the origin meets a line in the point 
(5, 2). What is the equation of the line ? 

5. The coordinates of the foot of the perpendicular dropped 
from the origin on a line are (a, b). Show that the equation 
of the line is 

ax + by = a 2 + b 2 . 

6. The line through the point (5, — 3) perpendicular to a 
given line meets it in the point (—3, 2). Find the equation 
of the given line. 

7. Prove that the line with intercepts 6 and 3 is perpen- 
dicular to the line with intercepts 3 and — 6. Is it also per- 
pendicular to the line with intercepts — 3 and 6 ? 



50 ANALYTIC GEOMETRY 

8. Prove that the line with intercepts a and b is perpen- 
dicular to the line with, intercepts b and — a. 

9. Show that the two points (5, 2) and (6, — 15) subtend a 
right angle at the origin. 

10. Prove that the two points, (x h y x ) and (x 2 , y 2 ), subtend a 
right angle at the origin when, and only when, x l x 2 -+- y x y 2 = 0. 

11. Do the points (6, —1) and (—3,4) subtend a right 
angle at the point (4, 6) ? At the point (- 4, - 2) ? 

12. Given the triangle whose sides lie along the lines, 

% — 2y + 6 = 0, 2x-y = S, x + y - 3 = 0. 

Find the coordinates of the vertices and the equations of the 
lines through the vertices parallel to the opposite sides. 

13. Two sides of a parallelogram lie along the lines, 

2x + 3y-6 = 0, ±x-y = L 

A vertex is at the point (—2, 1). Find the equations of the 
other two sides (produced). 

14. One side of a rectangle lies along the line, 

5x + 4?/- 9 = 0. 

A vertex on this side is at the point (1, 1) and a second vertex 
is at (2, —1). Find the equations of the other three sides 
(produced). 

15. For what value of A will the two lines, 

3a-22/ + 6 = 0, Atf-2/ + 2 = 0, 
(a) be parallel ? (b) be perpendicular ? 

16. For what value, or values, of m will the two lines, 

4cc — my + 6 = 0, x + my -f- 3 = 0, 
(a) be parallel ? (£) be perpendicular ? 

17. For what value of m will the two equations, 

mx + y -f- 5 = 0, 4cc + my + 10 = 0, 
represent the same line ? 



THE STRAIGHT LINE 51 

18. For what pairs of values for k and I will the two 
equations, 

12 x + ky + I = 0, Ix - 5 y + 3 = 0, 

represent the same line ? 

19. The equations of the sides of a convex quadrilateral are 

x = 2, y = 4, y = x, 2y = x. 

Find the coordinates of the vertices and the equations of the 
diagonals. 

20. Find the equation of the line through the point of 
intersection of the lines, 

- 3a; -5?/ -11 = 0, 2x -7y = ll, 

and having the intercept — 5 on the axis of y. 

21. Find the equation of the line through the point of in- 
tersection of the lines, 

2x + 5y = 4, Sx - ±y + 17 = 0, 

and perpendicular to the first of these two lines. 

22. Find the distance between the two parallel lines, 

Sx - ±y + 1 = 0, 6x - Sy + 9 = 0. 

Suggestion. Find the distance of a chosen point of the first 
line from the second. 

23. Let 

Ax + By+ (7=0 and Ax + By -f C = 

be any two parallel lines. Show that the distance between 
them is 

1^ C L or ± C '~ C . 



24. There are two points on the axis of x which are at the 
distance 4 from the line 2x — Sy — 4 = 0. What are their 
coordinates ? 

25. Find the coordinates of the point on the axis of y which 
is equidistant from the two points (3, 8), (—2, 5). 



52 ANALYTIC GEOMETRY 

26. There are two lines through the point (1, 1), each 
cutting from the first quadrant a triangle whose area is 2\. 
Find their slopes. Arts. — i — 2. 

27. Find the equation of the line through the point (3, 7) 
such that this point bisects the portion of the line between 
the axes. Ans. Ix + Sy — 42 = 0. 

28. The origin lies on a certain line and is the mid-point of 
that portion of the line intercepted between the two lines, 

Sx — 5y = 6, 4:X + y + 6 = 0. 

Find the equation of the line. Ans. x + 6y = 0. 

29. The line 

(1) 3a?-8y + 5 = 

goes through the point (1, 1). Find the equation of the line 

(2) through this same point, if the angle from the line (1) to 
the line (2) is 45°. Ans. llx-5y-6 = 0. 

30. Find the equations of the two lines through the origin 
making with the line 2x — 3 y = angles of 60°. 



CHAPTER III 
APPLICATIONS 

1. Certain General Methods. Lines through a Point. In 
many theorems and problems of Plane Geometry the question 
is to show that three lines pass through a point. Plane Geom- 
etry affords, however, no general method for dealing with this 
question. Each new problem must be discussed as if it were 
the first of this class to be considered. 

Analytic Geometry, on the other hand, affords a universal 
method, whereby in any given case the question can be settled. 
For, from the data of the problem, the equation of each of the 
lines can be found. These will all be linear, and can be writ- 
ten in the form 

L x : A x x + B iy + d = 0, 

L 2 : A 2 x + B 2 y + C 2 = 0, 

L 3 : A 3 x + B 3 y + C 3 = 0- 

The coordinates of the point of intersection of two of these 
lines, as i x and L 2 , can be found by solving the corresponding 
equations, regarded as simultaneous, for the unknown quanti- 
ties x and y. Let the solution be written as 

x = x', y = y'. 

The third line, L 3} will pass through this point (a/, y'), if and 
only if the coordinates of the latter satisfy the equation of L 3 ; 
i.e. if and only if 

A 3 x' + B 3 y' -f C 3 = 0. 

Points on a Line. A second question which presents itself 
in problems of Plane Geometry is to determine when three 

53 



54 



ANALYTIC GEOMETRY 



points lie on a straight line. Here, again, the repty of Analytic 
Geometry is methodical and universal. From the data of the 
problem it will be possible in any given case to obtain the 
coordinates of the three points. Call them 

(x 1} yi), (x 2y y 2 ), (x 3 , y s ). 

Now, we know how to write down the equation of a line 
through two of them, as (x ± , y^ and (x 2 , y 2 ). This equation 
will always be linear, and can be written in the form 

Ax + By + C=0. 

The third point, (x 3 ,y 3 ), will lie on this line if and only if its 
coordinates satisfy the equation of the line; i.e. if and only if 

Ax 3 + By 3 + C=0. 

The student should test his understanding of the foregoing 
theory by working Exs. 1-6 at the end of the chapter. 

2. The Medians of a Triangle. We recall the proposition 
from Plane Geometry, that the medians of a triangle meet in a 
point. The proof there 
given is simple, provided 
one remembers the con- 
struction lines it is desir- 
able to draw. By means, 
however, of Analytic Ge- 
ometry we can establish 
the proposition, not by 
artifices, but by the natural 
and direct application of 
the general principle enun- 
ciated in the preceding 
paragraph. 

The first step consists in 
the choice of the coordinate axes. This choice is wholly in 
our hands, and we make it in such a way as to simplify the 
coordinates of the given points. Thus, clearly, it will be well 




-B:(-l,3) 



£:tt,0) 



APPLICATIONS 55 

to take one of the axes along a side of the triangle. Let this 
be the axis of x. 

A good choice for the axis of y will be one in which this 
axis passes through a vertex. Let this be the vertex not on 
the axis of x. 

We begin with a numerical case, choosing the vertices A, _B, 
G at the points indicated in the figure. 

The Equations of the Medians. Consider the median AA'. 
One point on this line is given, namely A : (— 2, 0). A second 
point is the mid-point A' of the line-segment BC. By Ch. I, 
§ 5, the coordinates of A' are (2, 3). 

The student can now solve for himself the problem of finding 
the equation of the line L x through A : (— 2, 0) and A f : (2, 3). 
The answer is, 
L x \ 3x-±y + 6 = 0. 

In a precisely similar way the coordinates of B' are found 
to be ( — 1, 3), and the equation of the median BB' is 

L 2 : 3a + 5y-12 = 0. 

Finally, the coordinates of G are (1, 0), and the equation of 
the median CG is 
L z : 6x -f- y — 6 = 0. 

TJie Point of Intersection of the Medians. The next step con- 
sists in finding the point in which two of the medians, as L x 
and L 2 , intersect. The coordinates of this point will be given 
by solving as simultaneous the equations of these lines : 
3z-4y + 6 = 0, 
Sx + 5y - 12 = 0. 
The solution is found to be : 

* = -£, 2/ = 2. 

And now the third median, L 3i will go through this point, 
(■f, 2), if the coordinates of the point satisfy the equation of L 3} 

6 a; -f- 2/ — 6 = 0. 



56 



ANALYTIC GEOMETRY 



On substituting for x in this equation the value f and for y 
the value 2, we are led to the equation 

6.2+2-6 = 0. 

This is a true equation, and hence the three lines L i} L 2 , and 
L z pass through the same point. 

Remark. It can be shown by the formulas of Ch. I, § 6, 
that the above point (-§-, 2) trisects each of the medians AA\ 
BB\ and CO. 

EXERCISES 

1. Taking the same triangle as before, choose the axis of x 
along the side AB, but take the axis of y through A. The 
coordinates of the vertices will then be : 

A -.(0,0); B:(6,0);. 0:(2,6). 

Prove the theorem for this triangle. 

2. The vertices of a triangle lie at the points (0, 0), (3, 0), 
(0, 9). Prove that the medians meet in a point. 

3. Continuation. The General Case. We now proceed to 
prove the theorem of the medians for any triangle, ABC. Let 




A:(a,0) 



CH^fi) o 

Fig. 2 



^.•(5,0) 



the axes be chosen as in the text of § 2. Then the coordinates 
of A will be (a, 0), where a may be any number whatever, 



APPLICATIONS 57 

positive, negative, or zero. The coordinates of B will be (b, 0), 
where b may be any number distinct from a : 

b =£ a, or a — b^pO* 

Finally, the coordinates of C can be written as (0, c), where c 
is any positive number. 

Xext, find the coordinates of A', B\ C They are as shown 
in the figure. 

The equation of L x is given by Ch. II, (I), where 

(x 1 ,y l ) = (a,0); O2, 2/2) = Q, |\ 

x —a y — 

It is: I = c "' 

a - — 

2 2 

or 

L x : ex -f (2 a — &)# = ac. 

The equation of L 2 can be worked out in a similar manner. 
But it is not necessary to repeat the steps, since interchanging 
the letters a and b interchanges the points A and B, and also 
A' and B'. Thus L x passes over into L 2 . Hence the equation 
of L 2 is : 

L 2 : ex + (26 — a)y = be. 

The line i 3 is determined by its intercepts, ±{a 4- b) and c ; 
by Ch. II, (IV), its equation is found to be : 

L 3 : 2 ex -f (a + &)?/ = (a + b)c. 

To find the coordinates of the point in which L x and L 2 in- 
tersect, solve as simultaneous the equation of L x and L 2 : 

ex + (2 « — b)y = ac, 
ex + (2 b — a)# = 6c. 



The result is : 



a + b c 



* The figure has been drawn for the case in which a is negative and b 
positive. 



58 ANALYTIC GEOMETRY 

Finally, to show that this point, ( a ~}~ , -\ lies on L s , sub- 

\ 3 3J 

stitute its coordinates in the equation of L 3 : 

2c^ + (« + 6)|=(« + 6)c 

Since this is a true equation, the point lies on the line, and we 
have proved the theorem that the medians of a triangle pass 
through a point. 

That this point trisects each median can be proved as in the 
special case of the preceding paragraph, by means of Ch. I, § 6. 
The details are left to the student. 

EXERCISE 

Prove the theorem of the medians by taking the coordinate 
axes as in the first exercise of the preceding paragraph. Here, 
the vertices are 

^:(0,0); 3:(o,0); 0:(b 9 e), 

where a may be any number not 0, b any number whatever, 
and c any positive number. Draw the figure, and write in the 
coordinates of each point used. 

4. The Altitudes of a Triangle. Another proposition of 
Plane Geometry is, that the perpendiculars dropped from the 
vertices of a triangle on the opposite sides meet in a point. 

The proof of the proposition by Analytic Geometry is direct 
and simple. Let us begin with a numerical case, taking the 
triangle of Fig. 1. One of the perpendiculars is, then, the 
axis of y, and so all that is necessary to show is that the other 
two meet on this axis, or that the cc-coordinate of their point 
of intersection is 0. 

The equation of the line BC can be written down at once in 
terms of its intercepts : 

f + f = l, or 3x + 2y=12. 
4 6 



APPLICATIONS 59 

The slope of this line is A = — f . The slope of any line 
perpendicular to it is A' = § . Hence the equation of L l9 the 
perpendicular which passes through the point A : (— 2, 0), is 

2/-0 = l(z + 2), 
or 

L x : 2x - 3y + 4 = 0. 

In a similar manner the student can obtain the equation of 
the perpendicular L 2 from B on the side AC. It is, 
L 2 : x + 3^—4 = 0. 

On computing the a>coordinate of the point in which 7^ and 
L 2 intersect, it is found that x = 0, and hence the proposition 
is established for this triangle. 

Remark. For use in a later problem it is necessary to 
know the exact point in which the perpendiculars meet. It is 
readily shown that this point is (0, -§). 

EXERCISES 

1. Prove the above proposition for the special triangle con- 
sidered, choosing the coordinate axes as in Ex. 1 of § 2. 

2. Prove the proposition for the triangle of Ex. 2, § 2. 

3. Prove the proposition for the general case, choosing the 
axes as in Pig. 2. First show that the equation of the perpen- 
dicular L x from A on BC is 

L x : bx — cy — ab, 

and that the equation of the perpendicular L 2 from B on AC is 

L 2 : ax — cy = ab. 

Then show that these lines intersect each other on the axis 
of y. 

4. Show that the point in which the perpendiculars in the 

preceding question meet is (0, V 

5. Prove the theorem of the altitudes, when the axes of 
coordinates are taken as in the exercise of § 3. 



60 ANALYTIC GEOMETRY 

5. The Perpendicular Bisectors of the Sides of a Triangle. 

It is shown in Plane Geometry that these lines meet in a 
point. Since the student is now in full possession of the 
method employed in Analytic Geometry for the proof of this 
theorem, he will find it altogether possible to work out that 
proof without further suggestion. Let him begin with the 
special triangle of Fig. 1. He will find that the equations of 
the perpendicular bisectors of the sides are the following : 

L 2 : x + 3y-$ = 0-, 

L 3 : x - 1 = 0. 

These lines are then shown to meet in the point (1, J). 

He can work further special examples corresponding to the 
exercises at the end of § 2 if this seems desirable. 

Finally, let him work out the proof for the general case, tak- 
ing the coordinate axes as in Fig. 2. The three lines will be 
found to have the equations 

L x : bx—cy = |(6 2 - c 2 ), 

L 2 : ax — cy — \ (a 2 — c 2 ), 

L 3 : x = |(a + b). 

They meet in the point 

(a + b ab + c 2N 



EXERCISE 

Give the proof when the axes of coordinates are taken as in 
the exercise of § 3. 

6. Three Points on a Line. The foregoing three propositions 
about triangles have led to three points, namely, the three 
points of intersection of the three lines in the various Cases. 
In the case of the special triangle of Fig. 1, these points are 

(|, 2) ; (0, 4) ; (1, I). 



APPLICATIONS 61 

These points lie on a straight line. Let the student try to 
prove this theorem by Plane Geometry. 

The proof by Analytic Geometry is given immediately as a 
direct application of the second of the general principles enun- 
ciated in the opening paragraph of the chapter. 

Write down the equation of the line through two of these 
points, — say, through the first and third. It is found to be : 

3a; -3y + 4 = 0. 

The coordinates of the second point, 

* = °> y = h 

are seen to satisfy this equation, and the proposition is proved. 

EXERCISES 

1. Prove the proposition for the general case (Fig. 2). The 
points have been found to be : 

fa -\-b c\ /a q6\ f a + b ab + c - \ 

2. On plotting the three points obtained in the special case 
discussed in the text it is observed that the line-segment de- 
termined by the extreme points is divided by the intermedi- 
ate point in the ratio of 1:2. Prove this analytically. Is it 
true in general ? 

EXERCISES ON CHAPTER III 

1. Prove that the three lines, 

2x-3y-5 = 0, Sx + 4y - 16 = 0, 4a - 23y + 7 = 0, 
go through a point. 

2. Prove that the three lines, 

ax + by = 1, bx + ay = 1, x — y — 0, 

go through a point. 

3. Prove that the three points (4, 1), (—1, — 9), and 
(2, — 3) lie on a line. 



62 ANALYTIC GEOMETRY 

4. Prove that the three points (a, 6), (6, a), and 
(— a, 2a 4- b) lie on a line. 

5. Find the condition that the three lines, 

bx + ay = 2ab, ax + by = a 2 + b 2 , Sx — 2y = 0, 
where a 2 is not equal to b' 2 , meet in a point. 

6. Find the condition that the three points (a, b), (b, a), 
and (2 a, — b), where a is not equal to b, lie on a line. 

Lines through a Point 

7. Show that the line drawn through the mid-points of the 
parallel sides of a trapezoid passes through the point of inter- 
section of the non-parallel sides. 

8. Show that, in a trapezoid, the diagonals and the line 
drawn through the mid-points of the parallel sides meet in a 
point. 

9. A right triangle has its vertices A, B, and in the points 
(4, 0), (0, 3), and (0, 0). The points A : (4, - 4) and B' : (- 3, 
3) are marked. Prove that the lines AB', BA f , and the per- 
pendicular from on the hypothenuse meet in a point. 

10. (Generalization of Ex. 9.) Given a right triangle ABO 
with the right angle at O. On the perpendicular to OA in 
the point A measure off the distance AA', equal to OA, in the 
direction away from the hypothenuse. In a similar fashion 
mark the point B' on the perpendicular to OB in B, so that 
BB' = OB. Prove that the lines AB', BA', and the perpen- 
dicular from O on the hypothenuse meet in a point. 

11. Let P be any point (a, a) of the line x — y = 0, other 
than the origin. Through P draw two lines, of arbitrary slopes 
A x and a 2 , intersecting the #-axis in A 1 and A 2 and the ?/-axis 
in Bi and B 2 respectively. Prove that the lines A^ and 
A 2 Bi will, in general, meet on the line x + y — 0. 

12. If on the three sides of a triangle as diagonals paral- 
lelograms, having their sides parallel to two given lines, are 



APPLICATIONS 63 

described, the other diagonals of the parallelograms meet in a 
point. 

Prove this theorem, when the given lines are the coordinate 
axes, and the triangle has as its vertices the points (1, 6), 
(4, 11), (9, 3). 

13. Prove the theorem of the preceding exercise, when the 
given lines are the axes, and the triangle has its vertices in the 
points (0, 0), (a, a), (6, c). 

Points on a Line 

14. Show that in the parallelogram ABCD the vertex D, 
the mid-point of the side AB, and a point of trisection of the 
diagonal AC lie on a line. 

15. Prove that the feet of the perpendiculars from the 
point (2, — 1) on the sides of the triangle with vertices in the 
points (0, 0), (3, 0), and (0, 1) lie on a line. 

16. Prove that the feet of the perpendiculars from the point 
(—1, 4) on the sides of the triangle with vertices in the points 
(2, 0), (- 3, 0), and (0, 4) lie on a line, 

17. Show that the feet of the perpendiculars from the point 

( 0, —J on the sides of the triangle with vertices in the points 

(a, 0), (b, 0), and (0, c) lie on a line. 

18. Let M be the point of intersection of two opposite sides 
of a quadrilateral, and N t the point of intersection of the other 
two sides. The mid-point of MN and the mid-points of the 
diagonals lie on a right line. 

Prove this proposition for the special case that the vertices 
of the quadrilateral are situated at the points (0, 0), (8, 0), (6, 4). 
(1, 6). 

19. Prove the proposition of Ex. 18 for the general case. 
Suggestion. Take the axis of x through M and N, the 

origin being at the mid-point. The equations of the sides can 
then be written in the form 



64 ANALYTIC GEOMETRY 

y = X^x -h), y = X 2 (# - h), 
y = X 3 (oj + h) y y = \i(x + ft). 

20. Let be the foot of the altitude from the vertex O of 
the triangle ABC on the side AB. Then the feet of the per- 
pendiculars from on the sides BC and AC and on the other 
two altitudes lie on a line. 

Prove this theorem for the triangle ABC with vertices in 
the points (1, 0), (- 4, 0), (0, 2). 

21. Prove the theorem of the preceding exercise for the 
triangle with vertices in the points * (a, 0), (6,0), (0, c). It 
will be found that 

ac 1 cCc \ / be 2 Wc 



2 + c 2 ' tf + cy' V^ + c 2 ' 6 2 -j-c 2 / 
cCb — abc \ ( ab 2 — abc \ 
^Tc 1 ' a' + c-f y>' + c"' 6 2 + c 2 / 

are the coordinates of the four points which are to lie on a 
line, and that 

c(a 4- b)x + (ab — c l )y == a&c 

is the equation of the line. 



CHAPTER IV 



THE CIRCLE 

1. Equation of the Circle. According to Ch. I, § 7, the 

equation of the circle whose center is at the origin, and whose 
radius is p, is 

(i) <*+»■-/* 

In a precisely similar manner, the equation of a circle with 
its center at an arbitrary point 
C: (a, /?) of the plane, the length 
of the radius being denoted by p, is 
found to be : 




(2) (x- a y + (y-py = P ->. 

Example. Find the equation of 
the circle whose center is at the Fig. l 

point (— -J, 0), and whose radius is f . 

Here, a = — f , (3 = 0, and p = f . Hence, from (2) : 

(* + t) 2 + 2/ 2 = f 
This equation can be simplified as follows : 

* 2 + f*+¥ + ^ = *> 

or, finally, 

3a 2 + 3y 2 -f 8a; + 4 = 0. 



EXERCISES 

Find the equations of the following circles, and reduce the 
results to their simplest form. Draw the figure each time. 

1. Center at (4, 6) ; radius, 3. 

Ans. x 2 + y 2 -8x- 12y + 43 = 0. 

2. Center at (0, - 2) ; radius, 2. Arts, x 2 + y 2 + 4 y = 0. 

65 



66 ANALYTIC GEOMETRY 

3. Center at (— 3, 0) ; radius, 3. 

4. Center at (2, — 4) ; radius, 8. 

5. Center at (0, -|) ; radius, -J. 

6. Center at (3, — 4) ; radius, 5. 

7. Center at ( — 5, 12) ; radius, 13, 

8. Center at (■£, — J) ; radius, 2. 

9. Center at ( — f , f ) ; radius, J^. 

10. Center at (a, 0) ; radius, a. 

11. Center at (0, a) ; radius, a. 

12. Center at (a, a) : radius, «V2. 

2. A Second Form of the Equation. Equation (2) of § 1 can 

be expanded as follows : 

x i + y i-2ax- 2/3y + a* -f- /? 2 - p 2 = 0. 
This equation is of the form 

(1) tfi + f- + Ax + By+C=0. 

Let us see whether, conversely, equation (1) always repre- 
sents a circle. 

Example 1. Determine the curve represented by "he 
equation 

(2) a; 2 + 2/ 2 + 2a - 6y + 6 == 0. 
We can rewrite this equation as follows : 

(a* + 2» ) + (^_62/ ) = -6. 

The first parenthesis becomes a perfect square if 1 is added ; 
the second, if 9 is added. To keep the equation true, these 
numbers must be added also to the right-hand side. Thus 

(rf + 2» + l) + foi-6y+-9)=-6 + l+.9, 

or 

(z + l) 2 +0/-3)2 = 4. 

This equation is precisely of the form (2), § 1, where 
a = — 1, ft = 3, p s= 2. It therefore represents a circle whose 
center is at (— 1, 3), and whose radius is 2. 



THE CIRCLE 67 

Example 2. What curve is represented by the equation 

(3) ^ + ^ + 1 = 0? 

It is clear that no point exists whose coordinates satisfy this 
equation. For, x 2 and y 1 can never be negative. Their least 
values are 0, — namely, for the origin, (0, 0), — and even for 
this point, the left-hand side of the equation has the value -f 1. 
Hence, there is no curve corresponding to equation (3). 

Example 3. Discuss the equation 

(4) x 1 -f f- + 2x - ±y + 5 = 0. 
Evidently, this equation can be written in the form : 

(5) (x+iy+(y-~2y = 0. 

The coordinates of the point (—1, 2) satisfy the equation. 
But, for any other point (x, y), at least one of the quantities, 
x + 1 and y — 2, is not zero, and the left-hand side of the equa- 
tion is positive. Thus the point (— 1, 2) is the only point 
whose coordinates satisfy the equation. Hence equation (4) 
represents a single point (— 1, 2). 

Remark. Equation (5) can be regarded as the limiting case 
of the equation 

when p approaches the limit 0. This equation represents a 
circle of radius p for all positive values of p. When p ap- 
proaches 0, the circle shrinks down toward the point (— 1, 2) 
as its limit. Accordingly, equation (5) is sometimes spoken 
of as representing a circle of zero radius or a mill circle. 

The General Case. It is now clear how to proceed in the 
general case, in order to determine what curve equation (1) 
represents. The equation can be written in the form : 

(a* + Ax + i^ 2 ) + G/ 2 + By + ±JB*)= - C+ \A> + \B\ 
or 

-4:0 



(•+fH"f)'-^ 



68 ANALYTIC GEOMETRY 

If the right-hand side is positive, i.e. if 

A 2 + B 2 -±C>0, 

then equation (1) represents a circle, whose center is at the 
point (— %A, — IjB) and whose radius is 



p = ±VA 2 + B 2 -±C. 

If, however, A 2 -f B 2 — 4 C = 0, then equation (1) represents 
just one point, namely, (—^A, — ^B), — or, if one prefers, a 
circle of zero radius or a null circle. 

Finally, when A 2 -f B 2 — 4 C < 0, there are no points whose 
coordinates satisfy (1). To sum up, then : 

Equation (1) represents a circle, a single point, or there is no 
point whose coordinates satisfy (1), according as the expression 

A 2 + B 2 -4:C 

is positive, zero, or negative. 

Consider, more generally, the equation 

(6) a(x i + f) +bx + cy + d = 0. 

If a = 0, but b and c are not both 0, the equation represents 
a straight line. 

If, however, a ^= 0, the equation can be divided through by 
a, and it thus takes on the form : 

^ + 2/2+^ + -2/ + -=0. 
a a a 

This is precisely the form of equation (1), and hence the above 
discussion is applicable to it. 

EXERCISES 

Determine what the following equations represent. Apply 
each time the method of completing the square and examining 
the right-hand side of the new equation. Do not merely sub- 
stitute numerical values in the formulas developed in the text. 



THE CIRCLE 



69 



14 N 
¥97* 



1. & 2 + 2/ 2 + 6<c-8?/ = 0. 

.dbis. A circle, radius 5, with center at (—3, 4). 

2. a 2 + ?/ 2 - 6u; + 4y + 13 = 0. Ans. The point (3, - 2). 

3. #- + ?/ ? -f2a; + 42/-t-6=:0. J.ns. No point whatever. 

4. 80 + ^— 10«+24y = 0. 

5. x 2 + y 2 — 7x = 5. 

6. a 2 -fy 2 - 6^ + 8*/ + 25 = 0. 

7. 49aj 2 + 49^-14a; + 28y + 5 = 0. 

^4ws. The point (^-, 

8. a? + y l +8y = 10 ? 

9. a,' 2 + # 2 :=2aa,\ 

10. x 2 +y 2 = 2ay. 

11. a 2 + 2/ 2 -6a£-2fo/ + 9a° = 0. 

12. a 2 + 2/ 2 + 4aa-8fo/ + 16& 2 = 0. 

13. a^+^ + 3 = 0. 

14. x 2 + y 2 — 2x + 4y+10 = 0. 

15. 3a; 2 + 32/ 2 -4.T + 22/4-7 = 0. 

16. 5a 2 + 5?/ 2 - 6a + 8?/ = 12. 

17. 3x 2 + Sy 2 — x + y = 6. 



3. Tangents. Let the circle 
(1) ^+2/ 2 = /0 2 

be given, and let P x : (a^, 2/1) be any point of this circle. To 
find the equation of the tangent 
at P x . 

The tangent at P 1 is, by Ele- 
mentary Geometry, perpendicular to 
the radius, OP v Hence its slope, 
A/, is the negative reciprocal of the 
slope, yi/xx, of OP x ; or 






&<*»&) 




Fig. 2 



70 ANALYTIC GEOMETRY 

We wish, therefore, to find the equation of the line which 
passes through the point (x l9 y x ) and has the slope A/ = — x x /y v 
By Ch. II, § 2, (II), the equation of this line is 

(2) „_ ft „_&(,_«,). 

This equation can be simplified by multiplying through by 
y x and transposing : 

(3) x& + y$ = x? + 2/i 2 - 

Now, the point (x i} y x ) is, by hypothesis, on the circle ; hence 
its coordinates satisfy the equation (1) of the circle : 

*i 2 + 2/i 2 = P 2 - 

The right-hand side of equation (3) can, therefore, be replaced 
by the simpler expression, p% 

We thus obtain, as the final form of the equation of the 
tangent, the following : 

(4) XjX + y x y = p\ 

In deducing this equation it was tacitly assumed that y x ^= 0, 
since otherwise we could not have divided by it in obtaining 
A/. The final formula, (4), is true, however, even when y l = 0, 
as can be directly verified. For, if y x = and x x = p, then (4) 
becomes 

px = p 2 or x = p, 

and this is the equation of the tangent in the point (p, 0). 
Similarly, when y x = and x x = — p. 

Any Circle. If the given circle is represented by the 
equation 

(5) {x ^ a y H y_ (3 y = ^ 

precisely the same reasoning can be applied. The equation 
of the tangent to (5) at the point P x : to, y x ) of that circle is 
thus found to be: 

(6) fa -a)(x- «) + (2/i - /?)(*/ - /?)= p^. 
The proof is left to the student as an exercise. 



THE CIRCLE 71 

If the equation of the circle is given in the form 

(7) a 2 + y* + Ax + By + C=0, 

or in the form (6), § 2, the equation can first be thrown into 
the form (5), and then the equation of the tangent is given 

by (6). 

Example. To find the equation of the tangent to the circle 

(8) 3x* -|- 3?/ 2 + 8a - 5y = 
at the origin. 

First, reduce the coefficients of the terms in x 2 and y 2 to 
unity : 

# 2 4-2/ 2 + fa-f2/ = 0. 

Next, complete the squares : 

Or (» + f)«+(y._|.)l = f£ 

Now, apply the theorem embodied in formula (6). Since 
®i = 0, 2/i = 0, a = -f, /? = f, 
we have f (a + |) - Uv - *) = tt> 

or 8# — 5y = 0, 

as the equation of the tangent to (8) at the origin. 

EXERCISES 

Find the equation to the tangent of each of the following 
circles at the given point. 

1. x 2 + y- = 25 at (-3,4). Ans. Sx- 4# + 25 = 0. 

2. # 2 + y" 1 = a' at (0, a). Ans. y = a. 

3. x 9 + y 9 = 49 at (-7, 0). 

4. (jb - 1) 2 + (y + 2)2 = 25 at (4, 2). ^tis. Sx + 4.y = 20. 

5. (a + 5)2+(?/ - 3)2 = 49 at (2, 3). 

6. x 2 + y°~ — 9x + lly = at the origin. 



72 ANALYTIC GEOMETRY 

7. 2 a* + 2y* - Sx - y = 11 at (- 1, 2). 

8. Find the intercepts on the axis of x made by the tangent 
at (- 5, 12) to 

x 2 + y 2 = 169. Ans. - 33f 

9. Find the area of the triangle cut from the first quadrant 
by the tangent at (1, 1) to 

Sx 2 + 3y* + 8x + 16y = 30. 

10. If the equation 

x 2 + y 1 + Ax + £?/ + C = 

represents a circle, and if the point (x^ y{) lies on the circle, 
show that the equation of the tangent at this point can be 
written in the form : 

(9) x x x + ya + A(x + x 1 ) + ^(y + y 1 )+C=0. 

Suggestion. Find the values of a, /?, and p for the circle, 
substitute them in (6), and simplify the result. 

11. Do Exs. 6 and 7, using formula (9), Ex. 11. 

12. The same for the tangent to the circle in Ex. 9. 

13. Show that, if Pi : (x ly y x ) is any point of the circle 

x 1 + y 2 + Ax + By + = 0, 

at which the tangent is not parallel to the axis of y, then the 
slope of the tangent at P x is 

2xj + A 

2yi + B' 

4. Circle through Three Points. It is shown in Elementary 
Geometry that a circle can be passed through any three points 
not lying in a straight line. 

If the points are (aj l5 2/ x ), (x 2 , y 2 ), and (a? 3 , y 3 ), and if the 
equation of the circle through them is written in the form 

x* + y* + Ax + By+C=0, 

then clearly the following three equations must hold : 



THE CIRCLE 73 

xf + yf + ^ + -%! + C= 0, 

Xf + 2/2 2 + ^2 + %2 + C= 0, 
*8 2 + </3 ? + ^3 + %3 + C= 0. 

We thus have three simultaneous linear equations for de- 
termining the three unknown coefficients A, B, C. 

Suppose, for example, that the given points are the 
following : 

(1, 1), (1, - 1), (- 2, 1). 

The equations can be thrown at once into the form 
A + B + C = -2, 
A-B+C = -2, 

-2A + B+C = -5. 

Solve two of these equations for two of the unknowns in 
terms of the third. Then substitute the values thus found in 
the third equation. Thus the third unknown is completely 
determined, and hence the other two unknowns can be found. 

Here, it is easy to solve the first two equations for A and B 
in terms of C. On subtracting the second equation from the 
first, we find : 

2JB = 0; hence B = 0. 

Then either of the first two equations gives for A the value : 

A = -C-2. 

Next, set for A and B in the third equation the values just 
found : 

2C+4+<7 = -5, C = -3. 

Hence, finally, 

A = 1, B = 0, C = - 3, 

and the equation of the desired circle is : 

x 2 + y l + x - 3 = 0. 

Check the result by substituting the coordinates of the 
given points successively in this last equation. They are 
found each time to satisfy the equation. 



74 ANALYTIC GEOMETRY 

The circle through the three given points has its center in 
the point (- |, 0). Its radius is of length V05 = 1.803. 

EXERCISES 

Find the equations of the circles through the following 
triples of points. Plot the points and draw the circles. 

1. (1, 0), (0, 1), the origin. Ans. x* + y 2 — x — y = 0. 

2. (1,1), (-1, -1),(1, -1). 

3. (5,10), (6,9), (-2,3). 

4. The vertices of the triangle of Ex. 15 at the end of 
Ch. Ill, p. 63. Show that the point (2, — 1) of that exercise 
lies on the circle. 

5. The same question for Ex.- 17, p. 63. Show that the 

point (0, — ] of that exercise lies on the circle. 

6. The vertices of the triangle of Ch. Ill, Fig. 1. Find the 
coordinates of the center and check by comparing them with 
those of the point of intersection of the perpendicular bisectors 
of the sides of the triangle, as determined in Ch. Ill, § 5. 

7. The same question for the triangle of Ch. Ill, Fig. 2. 
Check. 

8. The vertices of the triangle formed by the coordinate 
axes and the line 2x — Sy = 6. 

9. The vertices of the triangle whose sides are : 

x-y-l = 0, £ + 2/4-2 = 0, 2a;-?/ + 3 = 0. 

Ans. Sx 2 + 3^ + 17a; + 16?/ + 25 = 0. 

EXERCISES ON CHAPTER IV 

1. Find the equation of the circle with the line-segment join- 
ing the two points (3, 0) and (5, 2) as a diameter. 

2. A circle goes through the origin and has intercepts — 5 
and 3 on the axes of x and y respectively. Find its equation. 

3. A circle goes through the origin and has intercepts a and 
b. Find its equation. 



THE CIRCLE 75 

4. Find the equation of the circle which has its center in 
the point (— 3, 4) and is tangent to the line Sx + Sy — 6 = 0. 

5. A circle has its center on the line 2x — 3y = and 
passes through the points (4, 3), (—2, 5). Find its equation. 

6. Find the equation of the circle which passes through the 
point (5, — 2) and is tangent to the line 3 x — y — 1 = at 
the point (1, 2). 

7. There are two circles passing through the points (3, 2), 
(—1.0) and having 6 as their radius. Find their equations. 

8. There are two circles with their centers on the line, 
ox — 3y = 8, and tangent to the coordinate axes. Find their 
equations. 

9. Find the equations of the circles tangent to the axes and 
passing through the point (1, 2). 

10. Find the equations of the circles passing through the 
points (3, 1), (1, 0) and tangent to the line x — y = 0. 

Suggestion. Demand that the center (a, /?) be equally 
distant from the two points and the line. 

11. Find the equations of the circles passing through the 
origin, tangent to the line x -f- y — 8 = 0, and having their 
centers on the line x == 2. 

12. Find the equations of the circles of the preceding exer- 
cise, if their centers lie on the line 2x — y — 2 = 0. 

13. Find the equation of the circle inscribed in the triangle 
formed by the axes and the line Sx — <iy — 12 = 0. 

14. Find the equation .of an arbitrary circle, referred to two 
perpendicular tangents as axes. 

15. Do the four points (0. 0), (6, 0), (0. - 4), (5, 1) lie on 
a circle ? 

16. Find the coordinates of the points of intersection of the 
circles 

x 2 + y 2 + 2x- y = 9. 



76 ANALYTIC GEOMETRY 

17. Find the coordinates of the points of intersection of the 

circles „ , 

a 2 + y 2 + aa + &2/ = 0, 

x 2 + y 2 + bx—ay = 0. 

Orthogonality 

18. A circle and a line intersect in a point P. The acute 
angle between the line and the tangent to the circle at P is 
known as the angle of intersection of the line and the circle at 
P. If the line meets the circle in two points, the angles of 
intersection at the two points are equal. Determine the angle 
in the case of the circle 

and the line 2x — y — 5 = 0. 

19. A circle and a line are said to intersect orthogonally if 
their angle of intersection is a right angle. Prove that the 
circle, 

x i + y i — ±x 4_ 6 ?/ + 3 = 0, 

is intersected by the line, 5x -f- y = 7, orthogonally. 

Suggestion. First answer geometrically the question : What 
lines cut a given circle orthogonally ? 

20. Show that the circle, 

tf+f + Jn + By +0=0, 

intersects the line, 

ax + by + c = 0, 

orthogonally when and only when 

aA + bB = 2c. 

21. If two circles intersect in a point P, the acute angle 
between their tangents at P is known as their angle of inter- 
section. If the circles intersect in two points, their angles of 
intersection at these points are equal. Find this angle in the 
case of the circles, 

a^ + ^ = 25, 
xi + y^-7x + y = Q. 



THE CIRCLE 77 

22. Prove geometrically that two circles intersect orthog- 
onally, that is, at right angles, when and only when the sum 
of the squares of their radii equals the square of the distance 
between their centers. Then show that the circles 

# 2 + y* — ±x + 5y- 2 = 0, 
2« 2 + 2f- + 4 a; - 6y - 19 = 0, 

intersect orthogonally. 

23. Prove that the two circles, 

X 2 + yi + AlX + Bl y + d = 0, 
x* + tf~ + A 2 x + £ 2 2/ + C 2 = 0, 
intersect orthogonally when and only when 
A ± A 2 + 5^2 = 2d + 2 2 . 

24. Find the equation of the circle which cuts the circle 

x 2 + y 2 + 2x = 

at right angles and passes through the points (1, 0) and (0, 1). 

25. There are an infinite number of circles cutting each of 
the two circles, 

x* + f - 4y + 2 = 0, 

x 2 + y 2 + 4?/ + 2 = 0, 

orthogonally. Show that they are all given by the equation 

x 2 + y* + ax-2 = 0, 

where a is an arbitrary constant. Where are their centers? 
Draw a figure. 

26. Find the equation of the circle cutting orthogonally the 
three circles, 

x* + yi = 9, 

a5 2 + 2/ 2 + 3 a j_5 2/ + 6== 0, 
cc 2 + yi - 2x + Sy - 19 = 0. 

Ans. x e, + y' i -\-10x + 9 = 0. 



78 ANALYTIC GEOMETRY 

Miscellaneous Theorems 

27. Prove analytically that every angle inscribed in a semi- 
circle is a right angle. 

28. Prove analytically that the perpendicular dropped from 
a point of a circle on a diameter is a mean proportional between 
the segments in which it divides the diameter. 

29. The tangents to a circle at two points P, Q meet in the 
point T. The lines joining P and Q to one extremity of the 
diameter parallel to PQ meet the perpendicular diameter in 
the points R and S. Prove that RT= ST. 

30. In a triangle the circle through the mid-points of the 
sides passes through the feet of the altitudes and also through 
the points halfway between the vertices and the point of inter- 
section of the altitudes. This circle is known as the Nine- 
Point Circle of the triangle. 

For the triangle with vertices in the points (— 4, 0), (2, 0), 
(0, 6) construct the circle and mark the nine points through 
which it passes. 

31. For the triangle in the preceding exercise find the equa- 
tion of the nine-point circle, as the circle through the mid- 
points of the sides. Ans. 3x 2 -f Sy 2 + Sx — 11 y = 0. 

32. Show that this circle goes through the other six points. 

33. For the triangle with vertices in the points (a, 0), (b, 0), 
(0, c) find the equation of the nine-point circle, as the circle 
through the mid-points of the sides. 

Ans. 2c(x l + y 2 )—(a + b)cx +(ab - c*)y = 0. 

34. Show that this circle goes through the other six points. 



CHAPTER V 

INTRODUCTORY PROBLEMS IN LOCI. SYMMETRY OF 
CURVES 

1. Locus Problems.* A point is moving under given condi- 
tions; its locus is required. This type of problem the student 
studied in Plane Geometry. But he found there no general 
method, by means of which he could always determine a locus ; 
for each problem he had to devise a method, depending on the 
particular conditions of the problem. 

Analytic Geometry, however, provides a general method for 
the determination of loci. Some simple examples of the 
method have already been given. Thus, in finding the equa- 
tion of a circle, we determined the locus of a point whose dis- 
tance from a fixed point is constant. Again, in deducing the 
equation of a line through two points, we found the locus of a 
point moving so that the line joining it to a given point has a 
given direction. 

The method in each of these cases consisted merely in ex- 
pressing in analytic terms — i.e. in the form of an equation 
involving the variable coordinates, x and y, of the moving 
point — the given geometric condition under which the point 
moved. We proceed to show how this method applies in less 
simple cases. 

Example 1. The base of a triangle is fixed, and the dis- 
tance from one end of the base to the mid-point of the opposite 
side is given. Find the locus of the vertex. 

* The locus problems in this chapter may be supplemented, if it is de- 
sired, by §§ 6-8 of the second chapter on loci, Ch. XIII, in which the loci 
of inequalities and the bisectors of the angles between two lines, together 
with related subjects, are considered. 

79 



80 



ANALYTIC GEOMETRY 



P:(x,V) 



Let the triangle be OAP, -with M as the mid-point of AP. 
Let a be the length of the base OA, and let I be the given 

distance. It is required to 
find the locus of P, so that 
always 
(1) OM= I. 

It is convenient to take 

x the origin of coordinates in 

and the positive axis of x 

along the base. The coordi- 

The coordinates of the moving 

The coordinates of the point 




M:(*±2:,V.\ 



A:(a,o) 



Fig. 1 



nates of A are then (a, 0). 
point P we denote by (x, y). 
M&Te 

'x + a 



The distance OM is 



\ 2 ' 2/ 



Thus condition (1), expressed analytically, is 

Z. 



WFW 



Squaring both sides of this equation and simplifying, we have 

(2) (x + ay+y* = (2l)\ 

This equation represents the circle whose center is at (—a, 0) 
and whose radius is 2 1. We have shown, therefore, that, if 
(1) is always satisfied, the coordinates (x, y) of P satisfy (2), 
and P lies on the circle. The locus of P appears, then, to be 
the circle. 

How do we know, though, that P traces the entire circle ? 
To prove this, we must show, conversely, that, if the coordi- 
nates (x, y) of P satisfy (2), condition (1) is valid. If . (x, y) 
satisfy (2), then, on dividing both sides of (2) by 4 and extract- 
ing the square root of each side, we obtain two equations : 



INTRODUCTORY PROBLEMS IN LOCI 



81 



i) 
ii) 



4 



+ «H 



x + a\ 

* J 



+ 






Equation ii) says that a positive or zero quantity equals a 
negative quantity, and is therefore impossible. Thus only 
equation i) remains. This equation says that OM= I. Hence 
condition (1) is satisfied by every point of the circle,* and so 
the circle is the locus of P. 

We have yet to describe the locus, independently of the 
coordinate system, with reference merely to the original tri- 
angle. _ Produce the base, in the direction from A to 0, to the 
point C, doubling its length. Then the locus of P is a circle, 
whose center is at C and whose radius is twice the given 
distance. 

Example 2. Determine the locus of a point P which moves 
so that the difference of the squares of its distances from two 
iixed points P 1; P 2 is constant, 
and equal to c : 



P:(x,y) 




Fig. 2 



[PP 1 2-PP 2 2=C, 

(3) 

\PP<?-PP? = C. 

Take the mid-point of the 
segment PiP 2 as origin and 
the axis of x along PiP 2 . 
The coordinates of P x and P 2 
can be written as (—a, 0), (a, 0) ; those of P, as (x, y). 

By Ch. I, § 3, 

PP,2 = (* + fc)2 + 2,2, P p 2 2 = (a . _ a y + y2 . 

Then the equations (3), expressed analytically, are 

* The two points in which the circle cuts the axis of x are exceptions, 
since these do not lead to a triangle, OAP. 



82 ANALYTIC GEOMETRY 

(x + a) 2 -f- f -(x - a) 2 -y 2 = c, 

(x - a) 2 -|- y 2 — (x + a) 2 — f = c. 
These reduce to 
(4) 4 «# = c, 4 a# = — c. 

Hence, if condition (3) is satisfied, P lies on one or the 
other of the lines 

c c 



(5) 



4a 4a 



Conversely, if P lies on one or the other of the lines (5), 
then (4) holds, and from (4) we show by retracing the steps 
that one or the other of the equations (3) is valid. 

Consequently, the locus of P consists of two straight lines, 
perpendicular to the line PiP 2 , and symmetrically situated 
with reference to the mid-point of PiP 2 , the distance of either 
line from the mid -point being c/4 a. Thus the locus consists 
of two entirely unconnected pieces, one corresponding to each 
of the equations (3). If c = 0, these equations are the same, 
and the two lines forming the locus coincide in the perpendic- 
ular bisector of the segment PiP 2 . 



EXERCISES 

In solving the following problems, the first step is to find 
the equation of a curve, — or the equations of curves, — on 
which points of the locus lie. The student must then take 
care (a) to show, conversely, that every point lying on the 
curve or curves obtained satisfies the given conditions ; and 
(6) to describe the locus, finally, without reference to the 
coordinate system used. 

1. A point P moves so that the sum of the squares of its 
distances to two fixed points P lf P 2 is a constant, c, greater 
than % PiP 2 2 . Show that the locus of P is a circle, with its 
center at the mid-point of PiP 2 . 

What is the locus if c = \ P, P 2 2 ? If c < \ P X PJ ? 



INTRODUCTORY PROBLEMS IN LOCI 83 

2. Find the locus of the mid-point of a line of fixed length 
which moves so that its end points always lie on two mutually 
perpendicular lines. 

3. Determine the locus of a point which moves so that the 
sum of the squares of its distances to the sides, or the sides 
produced, of a given square is constant. Is there any restric- 
tion necessary on the value of the constant ? 

4. Determine the locus of a point which moves so that the 
square of its distance to the origin equals the sum of its 
coordinates. Ans. A circle, center at (J, \), radius = \ V2. 

5. Show that the locus of a point which moves so that the 
sum of its. distances to two mutually perpendicular lines 
equals the square of its distance to their point of intersection 
consists of the arcs of four circles, forming a continuous curve. 
Where are the circles, and which of their arcs belong to the 
locus ? 

6. The base of a triangle is fixed, and the trigonometric 
tangent of one base angle is a constant multiple, not unity, 
of the trigonometric tangent of the other. Find the locus of 
the vertex. 

2. Symmetry. In the problems of the preceding paragraph, 
the equations of the loci were familiar and the curves they 
represented were easily identified. In subsequent chapters, 
however, we shall have locus problems to consider in which 
the resulting equations will be new 
to us. In drawing the curves which 

these equations represent, it will be <-*»»> 

useful to have at hand the salient 

facts concerning the symmetry of 
curves. 

Symmetry in a Line. Two points, 
P and P', are said to be symmetric 
in a line L, if L is the perpendicular bisector of PP'. 

If L is the axis of x and (x, y) are the coordinates of P, then 
it is clear that (x, — y) are the coordinates of P'. 



<(X,V) 



84 ANALYTIC GEOMETRY 

Similarly, if L is the axis of y and P has the coordinates 
(x, y), then P' has the coordinates (— x, y). 

Example 1. Given the curve 

(1) 2/2 = x. 

Let P : (x lf yi) be any point on it, i.e. let 

(2) . y 1 * = x l 

be a true equation. Then the point P' : (x h — y^, symmetric 
to P in the axis of x, also lies on the curve. 
' For, if we substitute the coordinates of P' into 

x i>y? (1), the result is ( — y x ) 2 = x h or (2), and (2) we 
know is a true equation. We say, then, that 

* the curve (1) is symmetric in the axis of x. 

The test for symmetry in the axis of x, 

ix v -vj} employed in this example, is general in appli- 

\. cation. We state it, and the corresponding 

Fig 4 test "^ or svmmetr y i n tne ax i s °f V> i n tne form 

of theorems. 
Theorem 1. A curve is symmetric in the axis of x if the sub- 
stitution of —y for y in its equation leaves the equation unchanged. 
Theorem 2. A curve is symmetric in the axis of y if the sub- 
stitution of — x for x in its equation leaves the equation unchanged. 

Symmetry in a Point. Two points, Pand P', are symmetric 
in a given point, if the given point is the mid-point of PP'. 

If the given point is the origin of 
coordinates and P has the coordinates 

(x, y), then the coordinates of P' are — 

evidently (- x, -y). (-*,-»°f 

Example 2. Consider the curve 5 

(3) y = x*. 

If P : (#!, 2/x) is any point on this curve, then the point 
P' : ( — x 1} — 2h), symmetric to P in the origin, is also on the 
curve. For, the condition that P' lies on the curve, namely, 

-2/i=(-^i) 3 ° r -2/i=-#i 3 , 



x 



INTRODUCTORY PROBLEMS IN LOCI 



85 





yp 



Fig. G 



is equivalent to the condition : y x = a^ 3 , that P lie on the curve. 
We say, then, that thp curve (3) is symmetric in the origin. 

This test, too, is general in application ; 
we formulate it as a theorem. 

Theorem 3. A curve is symmetric in the 
origin of coordinates, if the substitution of 
— x for x, and of — y for y, in it* equation 
leaves the equation essentially unchanged. 

A case in which the test leaves the 
equation wholly unchanged is that of the 
circle, x 2 -f- y 1 = p 2 , or the curve xy = a 2 
(Fig- 7). 

Now the circle in question is symmetric in both axes. It 
follows then, without further investigation, that it is sym- 
metric in the origin, the point of intersection of the axes. 
This conclusion holds always ; in fact, 
we may state the theorem. 

Theorem 4. If a curve is symmetric 
in both axes of coordinates, it is symmetric 
in the origin. 

The details of the proof are left to 
the student as an exercise. It is to be 
noted that the converse of the theorem, 
namely, that if a curve is symmetric in 
the origin, it is symmetric in the axes, is 
not true. For, the curve of Example 2 
is symmetric in the origin, but not symmetric in either axis ; 
this is true also of the curve xy == a 2 of Fig. 7. 



Fig. 7 



EXERCISES 

1. Prove Theorem 4. 

2. Test, for symmetry in each axis and in the origin, the 
curves given in the following exercises of Ch. I, § 7 : 

(a) Exercise 2 ; (c) Exercise 7 ; 

(6) Exercise 6 ; (d) Exercise 8. 



86 ANALYTIC GEOMETRY 

In each of the following exercises test the given curve for 
symmetry in each axis and in the origin. Plot the curve. 

3. xy + 1 = 0. 6. y 1 + 4 x = 0. 

4. 10?/ = a; 4 . 7. oj 2 — 2/ 2 = 4. 

5. 20a = 2/5. 8 . x 2 + 2?/ 2 = 16. 

EXERCISES ON CHAPTER V 

1. The base of a triangle is fixed and the ratio of the 
lengths of the two sides is constant. Find the locus of the 
vertex. Ans. A circle, except for one value of the constant. 

2. A point P moves so that its distance from a given line 
L is proportional to the square of its distance to a given point 
K, not on L. If P remains always on the same side of L as 
K, show that its locus is a circle. 

3. Find the locus of P in the preceding exercise, if it re- 
mains always on the opposite side of L from K. Does your 
answer cover all cases ? 

4. If, in Ex. 2, K lies on L and P may be on either side of 
L, what is the locus of P? 

5. Three vertices of a quadrilateral are fixed. Find the 
locus of the fourth, if the area of the quadrilateral is constant. 

6. Find the locus of a point moving so that the sum of 
the squares of its distances from the sides of an equilateral 
triangle is constant. Discuss all cases. 

Ans. A circle, center at the point of intersection of the me- 
dians ; this point ; or no locus. 

7. The feet of the perpendiculars from the point P: (X, Y) 
on the sides of the triangle with vertices in the points (0, 0), 
(3, 0), (0, 1) lie on a line. Find the locus of P. 

Ans. The circle circumscribing the triangle. 

8. The preceding problem, if the triangle has the points 
(2, 0), (- 3, 0), (0, 4) as vertices. 

9. Problem 7, for the general triangle, with vertices at 
(a, 0), (6, 0), (0, c). 



INTRODUCTORY PROBLEMS IN LOCI 87 

10. Show that the equation of the circle described on the 
line-segment joining the points (i\, y^, (x 2 , y 2 ) as a diameter 
may be written in the form 

(x - x{)(x - x 2 ) + (y- y x ){y - y 2 ) = 0. 

Suggestion. Find the locus of a point P moving so that the 
two given points always subtend at P a right angle. 

11. The two points, P and P', are symmetric in the line, 
x — y = 0, bisecting the angle between the positive axes of x 
and y. Show that, if (x, y) are the coordinates of P, then 
(y, x) are the coordinates of P'. 

12. Prove that a curve is symmetric in the line x — y = 0" 
if the interchange of x and y in its equation leaves the equation 
unchanged. 

13. If P and P' are symmetric in the line x -f- y = and 
P has the coordinates (x, y), show that the coordinates of P' 
are (-y, - x). 

14. Give a test for the symmetry of a curve in the line 
x + y=0. 

15. Test each of the following curves for symmetry in the 
lines x — y = and x -\-y = 0. 

(a) xy = a 2 ; (c) x l — y 2 = a 2 ; 

(b) xy = - a} ; (d) (x - yf- 2x-2y = 0. 

16. Plot the curve of Ex. 15, (d). 

In each of the following exercises find the equation of the 
locus of the point P. Plot the locus from the equation, mak- 
ing all the use possible of the theory of symmetry. 

17. The distance of P from the line x — 2 = equals its 
distance from the point (2, 0). 

18. The sum of the distances of P from the points (3, 0) 
and (- 3, 0) is 10. 

19. The difference of the distances of P from the points 
(5, 0) and f-5, 0) is 8. 



CHAPTER VI 



THE PARABOLA 



1. Definition. A parabola is defined as the locus of a point 
P, whose distance from a fixed line D is always equal to its 
distance from a fixed point F, not on 
the line. It is understood, of course, 
that P is restricted to the plane deter- 
mined by D and F. 

One point of the locus is the mid-point 
A, Fig. 2, of the perpendicular FE 
dropped from F on D. Through A 
draw T parallel to D. Then no other 
point on T, or to the left of T, can belong- 
to the locus, for all such points are 
clearly nearer to D than they are to F. 
the locus can be 
To the right of T 
D, cuttings AF, pro- 
in S. With ES as 
describe a circle, 




Further points of 
obtained as follows, 
draw L parallel to 
duced if necessary, 
radius and F as center 



cutting L in P and Q. Then P and Q 
lie on the locus. 

A large number of points having been 



obtained in this 



way, 



smooth curve 



i F 



Fig. 2 



can be passed through them. The curve 
is symmetric in the line AF, and evi- 
dently has rasa tangent. 

The line D is called the directrix, and the point F, the focus, 
of the parabola; A is the vertex, and the indefinite line AF, 
the axis ; FP is a focal radius. 

88 



THE PARABOLA 



89 



The student is familiar with, the fact that all circles are 
similar; i.e. have the same shape, and differ only in size. A 
like relation holds for any two parabolas. Think of them as 
lying in different planes, and choose in each plane as the unit 
length the distance be- 
tween the focus and the 
directrix. Then the one 
parabola, in its plane, is 
the replica of the other, 
in its plane. Conse- 
quently, the two parab- 
olas differ only in the 
scale to which they are 
drawn, and are, there- 
fore, similar. 

The details of the 
proof just outlined can be supplied at once by showing that 
the triangles FPM and MFE are similar, respectively, to 
F'PM' and M'F'E', the angles if/ in Fig. 3 being equal by 
construction. Hence 

FP = EF 
F'P' E'F'' 

i.e. focal radii, FP and F'P', which make the same angle with 
the axes always bear to each other the same fixed ratio. 




Fig 



EXERCISES 

1. Take a sheet of squared paper and mark D along one of 
the vertical rulings near the edge of the paper. Choose F 
at a distance of 1 cm. from D. Then the points of the locus 
on the vertical rulings — or on as many of them as -one desires 
— can be marked off rapidly with the compasses. Make a 
clean, neat figure. 

' 2. Place a card under the curve of Ex. 1 and, with a needle, 
prick numerous points of the curve through on the card, and 
mark, also, the focus and axis in this way. Cut the card alon.q 



90 



ANALYTIC GEOMETRY 



the curve with, sharp scissors. The piece whose edge is con- 
vex forms a convenient parabolic ruler, or templet, to be used 
whenever an accurate drawing is desired. 

A small hole at the focus and a second hole farther along 
the axis make it possible, in using the templet, to mark the 
focus and draw the axis. 

A second templet, to twice the above scale, will also be 
found nseful. 

3. The focus of a parabola is distant 5 units from the 
directrix. In a second parabola, this distance is 2 units. How 
much larger is the first parabola than the second, i.e., how do 
their scales compare with each other ? 



2. Equation of the Parabola. The first step is to choose 
the axes of coordinates in a convenient manner. Evidently, 

one good choice would be to take 
the axis of x perpendicular to D 
and passing through F. Let us do 
this, choosing the positive sense 
from A toward F. 

For the axis of y three simple 
choices present themselves, namely : 

(a) through A ; 

(b) along D ; 

(c) through F. 

Perhaps (b) seems most natural ; 

but (a) has the advantage that the 

curve then passes through the 

origin, and this choice turns out in practice to be the most 

useful one. We will begin with it. 

Let P : (x, y) be any point on the curve. Denote the dis- 
tance of F from D by m. Then 




Fig. 4 



EF 



m. 



AW-*, 



and 



BA-Z. 



THE PARABOLA 91 

ByCh. I, §3 

On the other hand, the distance of P from D is 
.By definition, these two distances are equal, or: 

/; 



(i) v(* 



+ ^ = x + -. 



Sqnare each side of the equation, so as to remove the radical, 
and expand the binomials : 

(2) a? — mx + — + y 2 = a 2 + ma 4- ^- 

The result can be reduced at once to the form 

(3) y' 1 = 2mx, 

and this is the equation of the parabola, referred to its vertex 
as origin and to its axis as the axis of x. 

The proof of this last statement is not yet, however, com- 
plete ; for it remains to show conversely that, if (x, y) be any 
point whose coordinates satisfy (3), it is a point of the parab- 
ola. From (3) we can pass to (2). On extracting the square 
root of each side of (2), we have two equations : 



■> V(-f) 



■ 2 , , , m 



■9 v(-iy+r— (' + f> 

one of which must be true, and both of which may conceivably 
be true. Now, x is a positive quantity or zero ; for, by 
hypothesis, the coordinates of the point (x, y) satisfy equation 
(3). Hence ii) is impossible, for it says that a positive or 
zero quantity is equal to a negative quantity. Thus only i) 



92 ANALYTIC GEOMETRY 

remains, and this equation is precisely the condition that the 
distance of (x, y) from D be equal to its distance from F. 
Hence the point (#, y) lies on the parabola, q. e. d. 



EXERCISES 

1. Show that the choice (&) leads to the equation 

(4) 7f = 2mx-m 2 . 

This is the equation of the parabola referred to its directrix 
and axis as the axes of y and x respectively, with the positive 
axis of x in the direction in which the curve opens. 

2. Show that the choice (c) leads to the equation 

(5) y 2 = 2mx + ra 2 . 

This is the equation of the parabola when the focus is the 
origin and the positive axis of x is along the axis of the curve 

in the direction in which the curve 

opens. 

P:(x y) 

3. Taking the axes as indicated in 
• x Fig. 5, show that the equation of the 
D parabola is 
Fig. 5 x 2 — 2my. 

4. Choosing the axis of y as in the foregoing question, show 
that the equation of the parabola is 

x 2 = 2my — m 2 , 
in case the axis of x is along D, and is 

x 2 = 2my + m 2 , 
in case F is taken as the origin. 

5. If the axis of x is taken along the axis of the parabola, 
but positively in the direction from F toward D, and if the 
origin is taken at the vertex, show that the equation of the 
curve is 

v 2 = — 2mx. 




THE PARABOLA 93 

6. If the axis of y is taken along the axis of the parabola^ 
but positively in the direction from F toward D, and if the 
origin is taken at the vertex, show that the equation of the 

curve is 

x 2 = — 2 my. 

7. Determine the focus and directrix of each of the follow- 
ing parabolas : 

(a) f< = 4<c. Ans. (1,0); x + 1 = 0. 

(b) y = x\ Ans. (0,1); 4^ + 1 = 0. 

(c) 3tf-5x=0. (d) 3y* + 22x = 0. 
(e) y = -2x*. (/) 5s 2 + 12^ = 
(g) y*=px. (h) x* = ±ay. 

8. It appears from the foregoing that any equation of the 
form 

y' l =z±Ax i or x* = ±Ay, 

where A is any positive constant, represents a parabola with 
its vertex at the origin. Formulate a general rule for ascer- 
taining the distance of the focus of such a parabola from the 
vertex. 

9. Find the equations of the following parabolas : 
(a) Vertex at (0, 0) and focus at (2, 0). 

(6) Vertex at (0, 0) and 2 x + 5 = as directrix. 

(c) Vertex at (0, 0) and focus at (0, - f ). 

(d) Vertex at (0, 0) and 2 y — 1 = as directrix. 

(e) Focus at (0, 0) and vertex at (— 3, 0). 

(/) Focus at (0, 0) and 3y + 4 = as directrix. 
(g) Focus at (6, 0) and axis of y as directrix. 
(h) Focus at (0, — 7) and axis of x as directrix. 

3. Tangents. The student will next turn to Chapter IX 
and study §§ 1, 2. It is there shown that the slope of the 
parabola 

(1) y- sc 2 mx 



94 ANALYTIC GEOMETRY 

at any one of its points (x l9 y x ) is, in general, given by the 
formula 

(2) A = ™; 

2/i 

and that the equation of the tangent line at any point (x 1} y{) 
can, without exception, be written in the form 

(3) yiy = m{x + Xi). 

Latus Rectum. The chord, PP', of a parabola which passes 
through the focus and is perpendicular to the axis is called the 
latus rectum (plural, latera recta). 

Its half-length is found by setting x = m/2 
in the equation of the parabola, and solving 
1 for the positive y : 

tf = 2m^=m\ y = m. 

Thus the length, PP, of the latus rectum 
is 2m. 

The tangent at either P or P' makes an 
angle of 45° with the axis of x. For, the 
slope of the tangent at P is, from (2) : 

m_m_ ^ 
2/i rn, 

Let E be the point in which the tangent at P meets the 
axis of x. Since FP = m, and AFEP=4:5 , EF = m and so 
E lies on the directrix. Consequently, the tangents at P and 
P* cut the axis of x at the point of intersection of the directrix 
with that axis. 

This theorem can also be proved by writing down the equa- 
tion of the tangent at P, 

. m 

and finding the intercept of this line on the axis of x. 



Fig. 6 



THE PARABOLA 95 

EXERCISES 

1. Find the equation of the tangent to the parabola y 1 — Sx 
at the point (12, 6). Arts, x - 4 y + 12 = 0. 

2. Find the equation of the normal to the same parabola 
at the given point. Ans. ±x + y = 54. 

3. Find the length of the latus rectum of the parabola of 
Ex.1. 

4. Show that the tangents to any parabola at the extremi- 
ties of the latus rectum are perpendicular to each other. 

5. Show that the tangent to the parabola y 1 = ±x at the 
point (36, 12) cuts the negative axis of x at a point whose dis- 
tance from the origin is 36. 

6. At what point of the parabola of Ex. 5 is the tangent 
perpendicular to the tangent mentioned in that exercise ? 

Ans. (^p -i). 

7. Show that the two tangents mentioned in Exs. 5 and 
6 intersect on the directrix, and that the chord of contact of 
these tangents, i.e. the right line drawn through the two points 
of tangency, passes through the focus. 

8. Show that the tangent to the parabola (1) at any point 
P cuts the negative axis of x at a point M whose distance 
from the origin is the same as the distance of P from the axis 
of y. 

9. Prove that the two parabolas, 

y- = 4x + 4 and ^ = _6a' + 9, 

intersect at right angles. Assume that the slope of the parab- 
ola of Ex. 2, § 2, at the point (x i} y x ) is m/y x . 

10. If two parabolas have a common focus and their axes 
lie along the same straight line, their vertices, however, being 
on opposite sides of the focus, show that the curves cut each 
other at right angles. 

4. Optical Property of the Parabola. If a polished reflector, 
like the reflector of the headlight of a locomotive or a search- 



96 



ANALYTIC GEOMETRY 



light, be made in the form of a paraboloid of revolution, i.e. 

the surface generated by a parabola which is revolved about 

its axis, and if a source of light be placed at the focus, the 

reflected rays will all be parallel. 

This phenomenon is due to the fact that the focal radius FP 

drawn to any point P of the parabola makes the same angle 

with the tangent at P as does the 
line through P parallel to the axis. 
The proof of this property can 
be given as follows. Let the 
tangent at P : (# b y L ) cut the axis 
of x in M. Then the length of 
OM is equal to x h by § 3, Ex. 8. 
Furthermore, OF = m/2. Hence 
the distance from M to F is 




Fig. 7 



MF=x x + 



But this is precisely the distance of P from D, § 2, and 
hence, by the definition of the parabola, it is also equal to FP. 
We have, then, that MF = FP. Consequently, the triangle 
MFP is isosceles, and 

y.FMP=y.MPF. 
But ^.FMP^^SPT, 

and the proposition is proved. 

The result can be restated in the following 

Theorem. The focal radius FP of a parabola at any point 
P of the curve and the parallel to the axis at P make equal angles 
with the tangent at P. 

Heat. If such a parabolic reflector as the one described 
above were turned toward the sun, the latter 's rays, being 
practically parallel to each other and to the axis of the reflector, 
would, after impinging on the polished surface, proceed along 
lines, all of which would pass through F. Thus, in particu- 
lar, the heat rays would be collected at F, and if a minute 



THE PARABOLA 97 

charge of gunpowder were placed at F, it might easily be 
fired. 

It is to this property that the focus (German, Brennpunkt) 
owes its name. The Latin word means hearth, or fireplace. 
The term was introduced into the science by the astronomer 
Kepler in 1604. 

EXERCISES ON CHAPTER VI 

1. A parabola opens out along the positive axis of y as axis. 
Its focus is in the point (0, 3) and the length of its latus 
rectum is 12. Find its equation. Ans. x 2 = 12 y. 

2. A parabola has its vertex in the origin and its axis along 
the axis of x. If it goes through the point (2, — 3), what is its 
equation ? Ans. 2y l — 9x = 0. 

3. Show that the equation of a parabola with the line x = c 
as directrix and with the point (c -f- ra, 0) or (c — ra, 0) as 
focus is 

y 2 = 2ra(a? — c) — ra 2 , or y 1 = — 2 m(x — c) — m 2 . 

Hence prove that every parabola with the axis of x as axis 
has an equation of the form : x — ay' 1 -+• b, where a and b are 
constants, a =£ 0. 

4. Find the equation of the parabola which has its axis 
along the axis of x and goes through the two points (3, 2), 
(- 2, - 1). Ans. Sx = htf- - 11. 

5. Prove that every parabola with an axis parallel to the 
axis of y has an equation of the form 

y — ax 2 -f- bx + c, 

where a, b, c are constants, a =£ 0. 

Suggestion. Find the equation of the parabola which has 
the line y = k as directrix and the point (I, k + m) or (I, k — m) 
as focus. 

6. Find the equation of the parabola which has a vertical 
axis and goes through the points (0, 0), (1, 0), and (3, 6). 

Ans. y = x* — x. 



98 ANALYTIC GEOMETRY 

7. A circle is tangent to the parabola y 1 = x at the point 
(4, 2) and goes through the vertex of the parabola. Find its 
equation. 

8. What is the equation of the circle which is tangent to 
the parabola y 1 = 2mx at both extremities of the latus rectum ? 

Ans. 4:X 2 + 4y • — 12 mx + m • = 0. 

9. Find the coordinates of the points of tangency of the 
tangents to the parabola y - = 2 mx which make the angles 60°, 
45°, and 30° with the axis of the parabola. Show that the 
abscissae of the three points are in geometric progression, and 
that this is true also of the ordinates. 

10. Show that the common chord of a parabola, and the 
circle whose center is in the vertex of the parabola and whose 
radius is equal to three halves the distance from the vertex to 
the focus, bisects the line-segment joining the vertex with the 
focus. 

11. Let N be the point in which the normal to a parabola 
at a point P, not the vertex, meets the axis. Prove that the 
projection on the axis of the line-segment PN is equal to one 
half the length of the latus rectum. 

12. On a parabola, P is any point other than the vertex, 
and N is the point in which the normal at P meets the axis. 
Show that P and N are equally distant from the focus. 

13. The tangent to a parabola at a point P, not the vertex, 
meets the directrix in the point L. Prove that the segment 
LP subtends a right angle at the focus. 

14. Show that the length of a focal chord of the parabola 
if = 2 mx is equal to x x + x 2 .+ m, where x Xl x 2 are the abscissae 
of the end-points of the chord. Hence show that the mid- 
point of a focal chord is at the same distance from the direc- 
trix as it is from the end-points of the chord. 

Exercises 15-26. The following exercises express properties 
of the parabola which involve an arbitrary point on the parab- 
ola. In order to prove these properties, it will, in general, be 



THE PARABOLA 99 

necessary to make actual use of the equation which expresses 
analytically the fact that the point lies on the parabola. 

15. An arbitrary point P of a parabola, not the vertex, is 
joined with the vertex A, and a second line is drawn through 
P, perpendicular to AP, meeting the axis in Q. Prove that 
the projection on the axis of PQ is equal to the length of the 
latus rectum. 

16. The tangent to a parabola at a point P, not the vertex, 
meets the tangent at the vertex in the point K. Show that 
the line joining K to the focus is perpendicular to the tangent 
at P. 

17. The tangent to a parabola at a point P, not the vertex, 
meets the directrix and the latus rectum produced in points 
which are equally distant from the focus. Prove this 
theorem. 

18. Prove that the coordinates of the point of intersection 
of the tangents to the parabola y*=2mx at the points (a^, yj, 
(%> 2/2) ma y De P u t iu the form 

\2m' 2 J 

Suggestion. To reduce the coordinates to the desired form, 
use the equations which express analytically the fact that the 
two points lie on the parabola. 

19. Show that the intercept on the axis of x of the line join- 
ing the points (x x , y 1 ) i (scg, y 2 ) of the parabola y 2 = 2mx may be 
expressed as 

2m 

By means of the results of the two preceding exercises prove 
the following theorems. 

20. The point of intersection of two tangents to a parabola 
and the point of intersection with the axis of the line joining 
their points of contact are equally distant from the tangent at 
the vertex, and are either on it or on opposite sides of it. 



100 ANALYTIC GEOMETRY 

21. Tangents to a parabola at the end-points of a focal 
chord meet at right angles on the directrix. 

22. If the points of contact of two tangents to a parabola 
are on the same side of the axis and at distances from the axis 
whose product is the square of half the length of the latus 
rectum, the tangents intersect on the latus rectum produced. 

23. The end-points of a chord of a parabola, which sub- 
tends a right angle at the vertex, are on opposite sides of the 
axis and at distances from the axis, whose product is the 
square of the length of the latus rectum. 

24. The chords of a parabola, which subtend a right angle 
at the vertex, pass through a common point on the axis ; this 
point is at a distance from the vertex equal to the latus rectum. 

25. The distance from the focus of a parabola to the point- 
of intersection of two tangents is a mean proportional between 
the focal radii to the points of tangency. 

26. The tangents to a parabola at the points P and Q inter- 
sect in T, and the normals at P and Q meet in N. Then the 
segment TM, where M is the mid-point of TN, subtends a 
right angle at the focus. 

Locus Problems 

27. Show that the locus of a point which moves so that 
the difference of the slopes of the lines joining it to two fixed 
points is constant is a parabola through the two fixed points 
What are its axis and vertex ? 

28. Determine the locus of a point which moves so that its 
distance from a fixed circle equals its distance from a fixed 
line passing through the center of the circle. 

Ans. Two equal parabolas, with foci at the center of the 
circle and axes perpendicular to the fixed line. 

29. The base of a triangle is fixed and the sum of the trigo- 
nometric tangents of the base angles is constant. Find the 
locus of the vertex. 



CHAPTER VII 

THE ELLIPSE 

1. Definition. An ellipse is defined as the locns of a point 
P, the sum of whose distances from two given points, F and 

F', is constant. It is found con- ___p 

venient to denote this constant 
by 2 a. Then 




Fig. 1 



(1) FP+F'P=2a. 

It is understood, of course, that P 
always lies in a fixed plane pass- 
ing through F and F'. 

The points F and F' are called 
the foci of the ellipse. It is clear that 2 a must be greater 
than the distance between them. 

Mechanical Construction. From the definition of the ellipse 
a simple mechanical construction readily presents itself. Let 
a string, of length 2 a, have its ends fastened at F and F', and 
let the string be kept taut by a pencil point at P. As the 
pencil moves, its point obviously traces out on the paper the 
ellipse. 

The student will find it convenient to use two thumb tacks 
partially inserted at F and F'. A silk thread can be tied to 
one of the thumb tacks and wound round the other so that 
it will not slip. Thus a variety of ellipses with different foci 
and different values of a can be drawn. 

Let the student make finally one ellipse in this manner, and 
draw it neatly. 

101 



102 



ANALYTIC GEOMETRY 



Center, Vertices, Axes. It is obvious from the definition, — 
and the fact becomes more striking from the mechanical con- 
struction, — that the ellipse is symmetric in the line through 
the foci. "" It is also symmetric in the perpendicular bisector 
of FF r . Hence it is symmetric, furthermore, in the mid-point, 

0, of the line FF'. 

The indefinite line through 
the foci, F and F', is called the 
transverse axis of the ellipse ; 
the perpendicular bisector of 
FF', the conjugate axis. The 
point is called the center of 
the ellipse; the points A, A', 
its vertices. 

The line-segments AA and 
BB', which measure the length and breadth of the ellipse, 
are known respectively as the major axis and the minor 
axis of the ellipse. The word "axes" refers sometimes to 
the transverse and conjugate axes, and sometimes to the major 
and minor axes, or their lengths, the context making clear in 
any case the meaning. 
When P is at A, equation (1) becomes 

FA + F'A = 2a. 
But FA = AF'. 




Hence 



AA'=2a 



and 



OA = a. 



Thus it appears that the length of the semi-axis major, OA, 
is a. Let the length of the minor axis be denoted by 26, and 
the distance between the foci by 2 c. Then, from the triangle 
FOB, we have : 
(2) a* = V + c\ 

Note that, of the three quantities a, b, and c, the quantity a 
is always the largest. 

Eccentricity. All circles have the same shape, i.e. are simi. 
lar ; and the same is true of parabolas. But it is not true of 



THE ELLIPSE 103 

ellipses. As a measure of the roundness or flatness of an 
ellipse a number, called the eccentricity, has been chosen ; this 
number is denned as the radio c/a and is denoted by e : 

(3) e=£. 

a 

Since c is always less than a, it is seen that the eccentricity 
of an ellipse is always less than unity : 

e<l. 

In terms of a and b, e has the value : 



(4) ,^, 

All ellipses with the same eccentricity are similar, and con- 
versely. For the shape of an ellipse depends only on b/a, the 
ratio of its breadth to its length, and since from (4) 



all ellipses for which the ratio b/a is the same have the same 
eccentricity, and conversely. 

A circle is the limiting case of an ellipse whose foci ap- 
proach each other, the length 2 a remaining constant. The 
eccentricity approaches 0, and a circle is often spoken of as an 
ellipse of eccentricity 0. 

EXERCISES 

1. The semi-axes of an ellipse are of lengths 3 cm. and 5 cm. 
Find the distance between the foci, and the eccentricity. 

Ans. 8; i. 

2. The eccentricity of an ellipse is § and the semi-axis 
minor is 4 in. long. How long is the major axis ? 

3. The major axis of an ellipse is twice as great as the 
minor axis. What is the eccentricity of the ellipse ? 

4. The major axis of an ellipse is 39 yards, and the eccen- 
tricity, 1 %. Find the minor axis. 



104 ANALYTIC GEOMETRY 

5. Express the eccentricity of an ellipse in terms of b and c. 

6. Show, from Fig. 2, that the eccentricity is given by the 
formula 

e = cos OFB. 

7. Give a proof, based on similar triangles, that two ellipses 
having the same eccentricity are similar. 

2. Geometrical Construction. Points on the ellipse may be 
obtained with speed and accuracy by a simple geometrical 

construction. Draw the major 

~^~ axis and mark the points A, F } 

F', A on it. Mark an arbitrary 

—+, h 1 — i — »— point Q between F and F. 

With F as center and AQ as 

> , radius describe a circle, and 

FlG 3 with F' as center and A'Q as 

radius describe a second circle. 

The points of intersection of these two circles will lie on the 

ellipse, since the sum of the radii is 

AQ + A'Q=2a. 

It is, of course, not necessary to draw the complete circles, 
but only so much of them as to determine their points of in- 
tersection. Moreover, four points, instead of two, can be ob- 
tained from each pair of settings of the compasses by simply 
reversing the roles of F and F f . 

EXERCISES 

1. Construct the ellipse for which c = 2± cm., a = 4 cm. 

2. From the ellipse just constructed make a templet, with 
holes at the foci and with the axes properly drawn. 

3. Construct the ellipse whose axes are 4 cm. and 6 cm. 

3. Equation of the Ellipse. It is natural to choose the axes 
of the ellipse as the coordinate axes (Fig. 4). Let the foci lie 



THE ELLIPSE 



105 



on the axis of x, and let P : (x, y) be any point of the ellipse. 
Then, from (1), § 1 . 

y _ 




Fig. 4 



(1) V(x-c) 2 +y 2 + V(x + c) 2 + y 2 = 2a. 
Transpose one of the radicals and square : 

(x — c) 2 + y 2 = (x + c) 2 + y 2 — 4 aV(# 4- c) 2 + y i + 4 a 2 . 
Hence 

(2) a . V(s 4- c) 2 + 2/ 2 = a 2 + ca. 
To remove this radical, square again : 

(3) a 2 x 2 + 2 cfe + a 2 c 2 + a 2 y 2 = a 4 + 2 a?cx + &x'\ 
or (a 2 - c 2 )a 2 + a?y 2 = a 2 (a 2 - c 2 ). 

But, by (2), § 1, 
and hence 

or 
(5) 



c 2 = 6 2 



6 2 ic 2 + a z y 2 = a 2 6 s , 



21 +£=1. 



This is the standard form of the equation of the ellipse, re- 
ferred to its axes as the axes of coordinates. The proof, how- 
ever, is not as yet complete, for it remains to show, conversely, 
that any point (x, y) whose coordinates satisfy equation (5) 
is a point of the ellipse. To do this, it is sufficient to show 
that x, y satisfy (1). From (5) we mount up to (4) and thence 
to (3), since all of these are equivalent equations. When. 



106 ANALYTIC GEOMETRY 

however, we extract a square root we obtain two equations 
each time, and so we are led, finally, to the four equations 



± V(a> - c) 2 + y 2 ± V(ar + c) 2 + f = '2 a, 

the ambiguous signs being chosen in all possible ways. The 
four equations can be characterized as follows : 

i) + +; ii) +.-; 

iii) - +; iv) - -. 

We wish to show that i) is the only possible one of the four 
equations. This is done as follows. 

Equation iv) is satisfied by no pair of values for x and y, 
since the left-hand side is always negative and so can never be 
equal to the positive quantity 2 a. 

Equations ii) and iii) say that the difference of the distances 
of (x, y) from F and F' is equal to 2 a, and hence greater than 
the line FF'= 2 c. Thus, in the triangle FPF r the difference 
of two sides is greater than the third side, and this is absurd.* 
Hence equations ii) and iii) are impossible and equation i) 
alone remains, q. e. d. 

Consequently, if we start with equation (5) as" given and 
require that a > b, then (5) represents an ellipse with 
semi-axes a and b and foci in the points (± c, 0), where 
c=Va 2 -6 2 . 

The Focal Radii. From equation (2) we obtain a simple 
expression for the length of the focal radius, FT. Dividing 
(2) by a and remembering that c/a = e, we have : 



V(a? + c) 1 4- y l = a + ex. 

But the value of the left-hand side of this equation is precisely 

F'P. Hence 

(6) F'P=a + ex. 

* If, in particular, the point (x, y) lay on FF', we should not, it is 
true, have a triangle. But it is at once obvious that in this case, too, 
equations ii) and iii) are impossible. 



THE ELLIPSE 107 

If, in transforming (1), the other radical had been transposed 
to the right-hand side and we had then proceeded as before, 
we should have found the equation : 



aw{x — c) ' + y l = a 2 — ex. 
From this we infer that 



V (x — c) 2 + y 2 = a — ex, 
or 
(7) FP = a-ex. 

EXERCISES 

1. What is the equation of the ellipse whose axes are of 

x 2 v 2 
lengths 6 cm. and 10 cm. ? Ans. — + ^- = 1. 

2. Find the coordinates of the foci of the ellipse of Ex. 1. 

3. The foci of an ellipse are at the points (1, 0) and (—1, 0), 
and the minor axis is of length 2. Find the equation of the 
ellipse. Ans. x 2 + 2y 2 = 2. 

4. Find the lengths of the axes, the coordinates of the foci, 
and the eccentricity of the ellipse 

25a; 2 + 1692/ 2 = 4225. 

5. An ellipse, whose axes are of lengths 8 and 10, has its 
center at the origin and its foci on the axis of y. Obtain its 
equation. 

6. Show that, if B > A, the equation 

A 2 B> 

still represents an ellipse with its axes lying along the axes of 
coordinates; but the foci lie on the axis of y at the points 
(0, C) and (0, - (T), where 

B 2 = A 2 + (P. 
The eccentricity is 

-i 



108 



ANALYTIC GEOMETRY 



7. Find the lengths of the axes, the coordinates of the foci, 
and the value of the eccentricity for each of the following 
ellipses : 



(a) 9« 2 + 42/ 2 = 36 

(b) 3a 2 + 21/2 = 12 

(c) x* + 2y 2 = 4 



(d) 5a? + 3^* = 46: 

(e) 2a 2 + 72/ 2 = 10 
(/) 11a 2 + 2/ 2 = 3. 



t' 
\f F J 



Fig. 5 



4. Tangents. The ellipse has the remarkable property that 
the tangent to the curve at any poiyit makes equal angles with 
the focal radii drawn to that point : 

i) Mechanical Proof. The simplest 

proof of this theorem is a mechanical 

one. Think of a flexible, inelastic string 

of length 2 a with its ends fastened at 

the foci, F and F'. Suppose a small, 

smooth bead to be threaded on this string. Let a cord be 

fastened to the bead and then pulled taut, so that the cord 

and the two portions of the string will be under tension. 

Evidently, the bead can be held in this manner at any point. 

(No force of gravity is supposed to act. The strings and bead 

may be thought of as resting on a smooth horizontal table.) 

The forces that act on the bead are : 

(a) the tension S in the cord ; 

(6) two equal tensions, R, in the string, 

directed respectively toward the foci. # 

Draw the parallelogram of forces for the 
forces B. It will be a rhombus, and so 
the resultant of these forces will bisect 
the angle between the focal radii. 

On the other hand, the force S, equal and opposite to this 
resultant, is perpendicular to the tangent at P. In fact, if 




Fig. 6 



* Since the bead is smooth, the tension in the string is the same at all 
its points, and so, in particular, is the same on the two sides of the bead. 



THE ELLIPSE 109 

instead of the flexible string we had a smooth rigid wire, in the 
form of the ellipse, for the bead to slide on, the bead would be 
held at P by the cord exactly as before. But the reaction of 
a smooth wire is at right angles to its tangent. This is the 
very conception of a smooth wire. For otherwise, if 8 were 
oblique, it could be resolved into a normal and a tangential 
component. But the smooth wire could not yield a reaction, 
part of which is along the tangent 

It follows, then, that the normal at P bisects the angle be- 
tween the focal radii, and hence these make equal angles with 
the tangent at P, q. e. d. 

ii) Proof by Means of Minimum Distances. A Lemma. A 
barnyard is bounded on one side by a straight river. The 
cows, as they come from the pasture, 
enter the barnyard by a gate at A, jV s 

go to the river to drink, and then > 

keep on to the door of the barn at B. — -p — 
What point, P, of the river should a " ^^ • , 

cow select, in order to save her steps FlG 7 

so far as possible ? 

It is easy to answer this question by means of a simple con- 
struction. From B drop a perpendicular BM on the line of 
the river bank, L, and produce it to B', making MB' — BM. 
Join A with B', and let AB' cut L at C. Then C is the posi- 
tion of P, for which the distance under consideration, 

^IP + PB, 

is least. 

For, the straight line AB' is shorter than any broken line 
APB' : 

AB'< APB'. 



But PB = PB' 
Hence 

AB'= AC + CB 


and CB = CB'. 

and APB'=AP+PB. 


It follows, then, that 




AC + CB<AP+PB, 



110 ANALYTIC GEOMETRY 

if P is any point of L distinct from C. Hence C is the point 
for which APB is a minimum. 

The point C is evidently characterized by the fact that 

-%ACN=y.BCM. 

We can state the result, then, by saying that the point P>for 
which the distance APB is least, is the point for ivhich 

%.APN=y.BPM. 

Optical Interpretation. We have used a homely example of 
cows and a barnyard. The problem we have solved is, however, 
identical with the optical problem of finding the point at 
which a ray of light, emanating from A, will strike a plane 
mirror L, if the reflected ray is to pass through B. For, the 
law of light is, that it will travel the distance in the shortest 
possible time, and hence it will choose the shortest path. 

Application to the Ellipse. The application of this result 
to the ellipse is as follows. The tangent to any smooth, closed, 
convex curve evidently is characterized 
by the fact that it meets the curve in 
one, and only one, point. 

Let P be any point of the ellipse. 
Draw the tangent, T, at P. Let Q be 
any point of T distinct from P. Now 

FlG ' 8 F'P + FP = F'R + FB, 

since the sum of the focal radii is the same for all points of an 
ellipse. But 

FR<BQ + FQ 
\ and so 

F'B + FR< F'B + BQ + FQ = F'Q + FQ. 

Therefore 

F'P+FP<F'Q + FQ. 

Hence P is that point of T for which the distance F'QF is 
least, and consequently the lines F'P and FP make equal 
angles with T, q. e. d. 




THE ELLIPSE 111 

EXERCISE 

Show that the normal of an ellipse at any point distinct from 
the vertices A, A' cnts the major axis at a point which lies 
between the foci. 

5. Optical and Acoustical Meaning of the Foci. Let a thin 
strip of metal, — say, a strip of brass a yard long and a quarter 
of an inch wide, — be bent into the form of an ellipse and 
polished on the concave side. Let a light be placed at one of 
the foci. Then the rays, after impinging on the metal, will 
be reflected and will come together again at the other focus, 
which will, therefore, be brilliantly illuminated. * 

The same is true of heat, since heat rays are reflected from 
a polished surface by the same law as that of light rays. If, 
then, a candle is placed at one focus and some gunpowder at the 
other, the powder can be ignited by the heat from the candle. 

Sound waves behave in a similar manner. The story is told 
of the Eatskeller in Bremen, the walls of which are shaped 
somewhat like an ellipse, that the city 
fathers were remarkably well informed 
concerning the feelings and views of the 
populace. For, the former drank their p IG . 9 

wine at a table which was situated at a 
focus, and thus could hear distinctly the conversation at a dis- 
tant table, which stood at the other focus and about which the 
Burger congregated. 

6. Slope and Equation of the Tangent. The student will 
next turn to Ch. IX, § 2, where the slope of the ellipse 

a 1 b- 

* The statement is, of course, strictly true only for such rays as travel 
in the plane through the foci, which is perpendicular to the elements of 
the cylinder formed by the polished hand. Since, however, only a nar- 
row strip of this cylinder is used, other rays will pass very near to the 
second focus and contribute to the illumination there. 




112 



ANALYTIC GEOMETRY 



at the point (x 1} y{) is found to be 



(1) 



V%. 



X = - 



a^i 



The equation of the tangent line at this point is 



(2) 






Lotus Rectum. The latus rectum of an ellipse is defined as 
a chord perpendicular to the major axis and passing through 
a focus. The term is also used to mean 
the length of such a chord. 
Thus, in the ellipse 

25 + 16 ' 
Fig. 10 

one focus is at the point (3, 0). The length 

of the latus rectum is twice that of the positive ordinate 
corresponding to this point. Setting, then, x = 3 in the equa- 
tion of the curve and solving for that ordinate, we have 




16 25 25' 



»-¥-* 



Hence the length of the latus rectum is 6$. 



EXERCISES 

1. Find the equation of the tangent to the ellipse 

225 25 
at the point (9, 4). Ans. x + ±y = 25. 

2. Find the equation of the normal to the ellipse of Ex. 1 
at the same point. Ans. 4a; — y = 32. 

3. At what point does the tangent to the ellipse 

2^ + 3^ = 14 
at the point (— 1, 2) cut the axis of y ? 



THE ELLIPSE 



113 



4. At what angle does the straight line through the origin, 
which bisects the angle between the positive axes of coordinates, 
cut the ellipse Sx 2 + ±y" = ~ ? Ans. 81° 53'. 

5. Find the area of the triangle cut off from the first quad- 
rant by the tangent to the ellipse of Ex. 3 at the point (1, 2). 

Ans. 8i 

6. Eind the length of the latus rectum of the ellipse of 
Ex. 1. Ans. 3J. 

7. The same for the ellipse of Ex. 3. 

8. Show that the length of the latus rectum of the ellipse 



£ = 1 

is given by any one of the expressions 



b<a, 



2a(l- 



Find its value in terms of c and e. 

9. Eind the length of the latus rectum of the ellipse 

25 x 2 + 16 y* = 400. Ans. 6|. 

10. Prove that the minor axis of an ellipse is a mean pro- 
portional between the major axis and the latus rectum. 



7. A New Locus Problem. Given a line D and a point F 
distant m from D. To find the locus of a point P such that 
the ratio of its distance FP from F to 
its distance MP from D is always 
equal to a given number, e : 

FP = 
MP 

It is understood that P shall be re- 
stricted to the plane determined by F 
and D. 

If, in particular, e = 1, the locus Fig. 11 



(i) 



or FP=€MF. 




114 ANALYTIC GEOMETRY 

is a parabola with D as directrix and F as focus ; Ch. 
VI, § 1. 

To treat the general case, let D be taken as the axis of y 
and let the positive axis of x pass through F. Then 



x. 



FP = -V{x- m) 2 + y\ MP = ± 

the lower sign holding only when x is negative, and (1) be- 
comes 



(2) -V(x - m) 2 + y 2 = ± ex. 

On squaring and transposing we obtain the equation : 

(3) (1 - e 2 )x 2 -2mx + tf + m 2 =0. 

This is the equation of the proposed locus. 
The student will now turn to Ch. XI and study carefully 
§ 1. 

EXERCISES 

1. Take e = \ and m = 3, the unit of length being 1 cm. 
With ruler and compasses construct a generous number of 
points of the locus, # and then draw in the locus with a clean, 
firm line. 

2. Work out the equation of the locus of Ex. 1 directly, 
using the method of the foregoing text, but not looking at the 
formulas. Ans. 3 x 2 + 4 y 2 — 24 x + 36 = 0. 

3. Take e = f and m = 4, the unit of length being 1 cm. 
Draw the locus accurately, as in Ex. 1. 

4. Work out directly the equation of the locus of Ex. 3. 

Ans. 16 x 2 + 25y 2 - 200 x = - 400. 

5. By means of a transformation to parallel axes show that 
the curve of Ex. 2 is an ellipse whose center is at the point 
(4, 0) and whose axes are of lengths 4 and 2V3. What is its 
eccentricity ? 

* The details of the construction are an obvious modification of the 
corresponding construction for the parabola in Ch. VI, § 1. A circle of 
arbitrary radius is drawn with its center at F, and this circle is cut by a 
parallel to D, whose distance from D is twice the radius of the circle. 



THE ELLIPSE 



115 



6. Show that the curve of Ex. 4 is an ellipse whose axes 
are 7-J and 6. What is its eccentricity ? 

8. Discussion of the Case e < 1. The Directrices. From 

equation (3) of § 7 follows: 



(1) 



, 2 m , v 2 
a,' 2 : X -f 



71V 



1 _ e 2 1 _ £ 2 1 _ £ 2 

The first two terms on the left-hand side are also the first two 
in the expansion of 

m \ 2 9 2 m . m 2 

) = x 2 x -\ 

1 - e 2 ; 1 - e 2 (1 - c 2 ) 2 

If, then, we add the third term of the last expression to both 
sides of (1), we shall have : 
2 m , m 2 



x'- — 



1-e 2 



x + 



+ 



y 1 m 2 



ni l 



(1-e 2 ) 2 1-e 2 (1-e 2 ) 2 1-e 2 ' 



or 
(2) 



X — 



m 



y 2 e 2 m 2 



1- 



(1 - e 2 ) 2 

This equation reminds us strongly of the equation of an 
ellipse. In fact, if we transform to parallel axes with the 
new origin, 0', at the point 

m 



the equations of transformation are 
(3) *_,__£_, 

and (2) then takes on the form 



^o = 0, 



(4) 

or 
(5) 

where 
(6) 



x' 2 



V 



l_ e 2 (1_ £ 2)2' 



a- o- 




Fig. 12 



b = 



em 



V1 - ? 



116 



ANALYTIC GEOMETRY 



Thus the locus is seen to be an ellipse with its center, 0', at 
the point 

(?) (^,0) 

the semi-axes being given by (6). 

The value of c is given by the equation c 2 = a 2 — b' 2 . Hence 



(8) 



c == 



1-e 2 

The eccentricity, e = c/a, is now seen to be precisely e : 

e = c; 
i.e. the given constant, e, turns out to be the eccentricity of the 



Finallv, F is one of the foci. For, the distance from F to 0' 



is 



OO f -OF = 



m 



em 



1 _ e 2 1 _ £ 2> 

and this, by (8), is precisely c. 

The line D is called a directrix of the ellipse. Its distance 
from the center is 

m me l__a 

"^"l-e^T"!' 



00' 



The Directrices. From the symmetry of the ellipse it is 
clear that there is a second directrix, D', on the other side of 

the conjugate axis, parallel to that 
axis, and at the same distance 
from it as D. This line D' and 
the focus F f stand in the same re- 
lation to the ellipse as the first 
D line, D, and the focus F. Thus 
the ellipse is the locus of a point 
so moving that its distance from a 
focus always bears to its distance from the corresponding 
directrix the same ratio, e, the eccentricity. 

Since the distance of D from the center of the ellipse is a/e, 
the equations of the directrices of the ellipse 




Fig. 13 



are 



THE ELLIPSE 117 



5+g= i > *>*>> 



X = — - # = -< 

e e 



EXERCISES 

1. Show that the distances of the vertices, A and A', from 
are : 

OA = -^-, OA' 



1 + e' 1-e 

2. Collect the foregoing results in a syllabus, arranged in 
tabular form, giving each of the quantities a, b, c, 00', OA, 
OA', OF, OF' in terms of m and e. 

3. Work out each of the quantities of Ex. 2 directly for the 
ellipse of § 7, Ex. 4, and verify the result by substituting the 
values e = f , m = 4 in the formulas of the syllabus. 

4. Between the jfrre constants of the ellipse, a, b, c, e, m, 
there exist three relations, which may be written in a variety 
of ways ; as, for example, 

i) cC- = 6' + c - ; ii) e = - ; iii) m = ~~ e a. 

a e 

By means of these relations, any three of the five quantities 
can be expressed in terms of the other two. Thus, in Ex. 2, 
m and e are chosen as the quantities in terms of which all 
others shall be expressed. 

Taking the semi-axes, a and b, (a>b), as the preferred pair, 
express the other quantities in terms of them. 

5. Show that the tangent to the ellipse 

25^16 

at an extremity of a latus rectum cuts the transverse axis in 
the same point in which this axis is cut by a directrix. 



118 



ANALYTIC GEOMETRY 



6. The same for any ellipse. 

7. Prove directly that, if P is any point of the ellipse 



a 2 V 



b < a, 



the ratio of its distance from a focus to its distance from the 
corresponding directrix is equal to the eccentricity. 

8. Show that in an ellipse the major axis is a mean propor- 
tional between the distance between the foci and the distance 
between the directrices. 

9. Show that the distances from the center and a focus of 
an ellipse to the directrix corresponding to the focus are in 
the same ratio as the squares of the semi-axis major and the 
semi-axis minor. 

9. The Parabola as the Limit of Ellipses. We have proved 
that, when e < 1, equation (3), § 7, represents an ellipse with 

eccentricity e — e. We 
know that , if e = 1, the 
equation represents a 
parabola. If, then, in the" 
equation we allow e to 
approach 1 through values 
< 1, the ellipse which the 
equation defines ap- 
proaches a parabola as its 
limit. 

We can visualize the 
ellipse, going over into 
a parabola, by drawing 
a number of ellipses 
having the same value of 
are increasing toward 1 
. The directrix D, 




Fig. 14 



m, but having values for c 
as their limit, viz. e == -J-, 



which 



along the axis of y, and the focus F: (m, 0) are the same for 
all the ellipses. But the center 0' and the right-hand vertex 



THE ELLIPSE 



119 



A' of each successive ellipse are farther away from 0. and 
their distances from 0, namely, 



00' 



m 



0A' = 



m 



1-e 2 ' 1-e' 

increase without limit. Thus, as e approaches 1, the ellipse 
approaches as its limit the parabola whose directrix is D and. 
whose focus is F. 

10. New Geometrical Construction for the Ellipse. Para- 
metric Representation. Let it be required to draw an ellipse 
when its axes, A A' and BB\ are given. 
Describe circles of radii a = OA and 
b = OB, with the origin O as the 
common center. Draw any ray from 
O, making an angle <j> with the posi- 
tive axis of x, as shown in the 
figure. Through the points Q and R 
draw the parallels indicated. Their 
point of intersection, P, will lie on 
the ellipse. For, if the coordinates of 
Pbe denoted by (#, y), it is clear that 

(1) x = a cos 4>, y = b sin <f>. 

From these equations <f> can be eliminated by means of the 
trigonometric identity 




Fig. 15 



Hence 
(2) 



sin 2 <f> -1- cos 2 <£ = 1. 



if- 



Conversely, any point (x, y) on the ellipse (2) has corre- 
sponding to it an angle <f>, for which equations (1) are true. 

Equations (1) afford what is known as a parametric repre- 
sentation of the coordinates of a variable point (x, y) of the 
ellipse in terms of the parameter <f>. When b = a, the ellipse 
becomes a circle, and the equations (1) become 

(3) x—a cos <j>, y = a sin <f>. 



120 



ANALYTIC GEOMETRY 




These parametric representations, though, little used in Ana- 
lytic Geometry, are an important aid in the Calculus. 

The larger of the two circles in Fig. 
15 is commonly called the auxiliary 
circle of the ellipse, and the points It 
and P are known as corresponding points. 
The angle <f> is called the eccentric angle. 

EXERCISE 

FlG By means of the foregoing method, 

draw on squared paper an ellipse whose 
axes are of length 4 cm. and 6 cm. 

EXERCISES ON CHAPTER VII 

1. The earth moves about the sun in an elliptic orbit.* The 
shortest and longest distances from it to the sun are in the 
ratio 29 : 30. What is the eccentricity of the orbit ? 

2. Show that the slopes of the tangents to an ellipse at the 
extremities of the latera recta are ± e. 

3. The axes of an ellipse which goes through the points 
(4, 1), (2, 2) are the axes of coordinates. Find its equation. 

4. The center of an ellipse is in the origin and the foci are 
on the axis of x. The ellipse has an eccentricity of -| and goes 
through the point (12, 4). What is its equation? 

25^16 25 

5. Solve the preceding problem if the foci may lie on either 
axis of coordinates. 

6. Find the equations of the ellipses which have the axes 
of coordinates as axes, go through the point (3, 4), and have 
their major and minor axes in the ratio 3 : 2. 

7. Show that the ellipses represented by the equation 

2x i + 3y*- = '&, 

* The planets describe ellipses about the sun as a focus, and the comets 
usually describe parabolas with the sun as the focus. 



THE ELLIPSE 121 

where c 2 is an arbitrary positive constant, are similar. What 
is the common value of the eccentricity ? 

8. How many ellipses are there with eccentricity -J-, having 
their centers in the origin and their foci on the axis of a:? 
Deduce an equation which represents them all. 

Ans. 3 x 2 + 4?/* = c 2 . 

9. The foci of an ellipse lie midway between the center 
and the vertices. What is the eccentricity ? How many such 
ellipses are there, with centers in the origin and foci on the 
axis of x? Write an equation which represents them all. 

10. The line joining the left-hand vertex of an ellipse with 
the upper extremity of the minor axis is parallel to the line 
joining the center with the upper extremity of the right- 
hand latus rectum. Answer the questions of the preceding 
exercise. 

11. The foci of an ellipse subtend a right angle at either 
extremity of the minor axis. What is the eccentricity? 
Find the equation of all such ellipses with centers in the 
origin and foci on the axis of y. 

12. Prove that the ratio of the distance from a focus of an 
ellipse to the intersection with the transverse axis of the 
normal at a point P, and the distance from this focus to P 
equals the eccentricity of the ellipse. 

13. The projections of a point P of an ellipse on the trans- 
verse and conjugate axes are P x and P 2 . The tangent at P 
meets these axes in T x and T 2 . Prove that OPj • OT x = a 2 and 
OP 2 ' OT 2 = b 2 , where is the center and a and b are the 
semi-axes of the ellipse. 

14. Prove that the segment of a tangent to an ellipse be- 
tween the point of contact and a directrix subtends a right 
angle at the corresponding focus. 

15. Determine the points of an ellipse at which the tangents 
have intercepts on the axes whose absolute values are propor- 
tional to the lengths of the axes. 



122 ANALYTIC GEOMETRY 

16. Through, a point M of the major axis of an ellipse a line 
is drawn parallel to the conjugate axis, meeting the ellipse in 
P and the tangent at an extremity of the latus rectum in Q. 
Show that the distance MQ equals the distance of P from the 
focus corresponding to the latus rectum taken. 

17. Prove that the line joining a point P of an ellipse with 
the center and the line through a focus perpendicular to the 
tangent at P meet on a directrix. 

' 18. Prove that the distance from a focus F to a point P of 
an ellipse equals the distance from F to the tangent to the 
auxiliary circle at the point corresponding to P. 

19. Find the equation of a circle which is tangent to the 
ellipse 

^4-^ = 1 
a* b 2 

at both ends of a latus rectum. 

20. In an ellipse whose major axis is twice the minor axis, 
a line of length equal to the minor axis has one end on the 
ellipse, the other on the conjugate axis. The two ends are 
always on opposite sides of the transverse axis. Prove that 
the mid-point of the line lies always on the transverse axis. 

21. A number of ellipses have the same major axis both in 
length and position. A tangent is drawn to each ellipse at 
the upper extremity of the right-hand latus rectum. Prove 
that these tangents all pass through a point. 

Exercises 22-28. In these exercises, in which properties in- 
volving an arbitrary point P of an ellipse are to be proved, it 
will, in general, be necessary to make actual use of the equa- 
tion expressing the fact that the point P lies on the ellipse. 

22. The tangent to an ellipse at a point P meets the tan- 
gent at one vertex in Q. Prove that the line joining the other 
vertex to P is parallel to the line joining the center to Q. 

23. The lines joining the extremities of the minor axis with 
a point P of an ellipse meet the transverse axis in the points 



THE ELLIPSE 123 

M and N. Prove that the semi-axis major is a mean propor= 
tional between the distances from the center to M and N, 

24. Prove the theorem of the preceding exercise when the 
major and minor axes, and the transverse and conjugate axes, 
are interchanged. 

25. Show that the segment of a directrix, between the 
points of intersection of the lines joining the vertices with a 
point on an ellipse, subtends a right angle at the correspond- 
ing focus. 

26. Prove that the product of the distances of the foci of 
an ellipse from a tangent is a constant, independent of the 
choice of the tangent. 

27. Let F' and F be the foci of an ellipse and P any point 
on it. Prove that b 2 : FK 1 = F'P: FP, where FK is the dis- 
tance from F to the tangent at P. 

28. The normal to an ellipse at a point P meets the axes in 
N x and JV 2 . Show that PN X • PN 2 is equal to the product of 
the focal radii to P. 

Loci 

29. A point moves so that the product of the slopes of the 
two lines joining it to two fixed points is a negative constant. 
What is its locus ? 

30. A circle whose diameter is 10 cm. is drawn, center at 0. 
On a radius OA a point B is marked distant 4 cm. from 0. 
If OQ is any second radius, show how to construct, with ruler 
and compasses, a point P on OQ, whose distance from the 
circle equals its distance from B. In this way plot a number 
of points on the locus of P. 

31. Find the equation of the locus of the point P of the 
preceding exercise. Take the origin of coordinates at the 
mid-point of OB. 

32. The base of a triangle is fixed and the product of the 
tangents of the base angles is a positive constant. Find the 
locus of the vertex. 



CHAPTER VIII 



THE HYPERBOLA 





Fig. 1 



1. Definition. A hyperbola is defined as the locus of a point 
, the difference of whose distances from two given points, F 

and F', is constant. It is found 
convenient to denote this constant 
by 2 a. Then 

FP-F'P=2a, 
or F'P-FP=2a. / 

It is understood, of course, that P 
is restricted to a particular plane 
through F and F'. 
The points F and F f are called the foci of the hyperbola. 
It is clear that 2 a must be less than the distance between 
them. Denote this distance by 2 c. 

Geometrical Construction. Draw the indefinite line FF', 
mark the mid-point, 0, of the segment FF' } and the points A 
and A' each at a distance a 
from 0: P + P * 

OA=OA' = a; OF=OF' = c. 

— h — b •- 

The point A lies on the locus ; F A ° 

for, ^T 

FA = c — a, F'A = c + a, 

and hence F'A - FA = 2 a. 

Likewise, u4' lies on the curve. 

Mark any point, N, to the right of F. With radius AN&nd 
center F, describe a circle. Next, with radius ^L'^Tand center 

124 



Q 



-t— 4- 



A f JV 



Q 



Fig. 2 



THE HYPERBOLA 



125 




F'A 



B' 
Fig. 3 




A\F 



F', describe a second circle. The points P and Q in which 
these circles intersect are points of the locus. For, 

F'P- FP = A'N- AN= A' A = 2 a. 

Two more points, P' and Q', can be obtained from the same 
pair of settings by interchanging the centers, F and F', of the 
circles. 

By repeating the construction a number of times, a goodly 
array of points of the hyperbola can be obtained. These 
points will lie on two distinct arcs, 
symmetric to each other in the 
perpendicular bisector BOB' of 
FF'. Thus it will be seen that 
the hyperbola consists of two 
parts, or branches, as they are 
called. These branches, besides 
being the images of each other in 
BB', are each the image of itself 

in FF'. It is natural to speak of the indefinite straight lines 
FF' and BB' as the axes of the hyperbola. FF' is called the 
transverse, BB' the conjugate axis ; O is the center, and A, A' 
are the vertices. 

EXERCISES 

1. Taking c = 3 cm. and a = 2 cm., make a clean drawing of 
the corresponding hyperbola. 

2. Reproduce the drawing on a rec- 
tangular card and, with a sharp knife 
or a small pair of scissors, cut out the 
center of the card along the hyperbola 
and two parallels to the transverse axis. 
On the templet which remains make 
holes at the foci and draw the two axes. 





Fig. 4 



2. Equation of the Hyperbola. The treatment here is paral- 
lel to that of the ellipse, Ch. VII, § 3. Let the transverse axis 



126 



ANALYTIC GEOMETRY 



be chosen as the axis of x ; the conjugate axis, as the axis of y. 
Then the equation of the right-hand branch of the hyperbola 
can be written in the form 

(x,v) (1) V(a> + c) 2 + f- - s/(x - c) 2 + y 2 = 2 a. 

( c ,o) Transpose the first radical and square : 

( x -c) 2 +y 2 =(x + c) 2 +y 2 




F:(-c,o), 



Fig. 5 



4aV(a + c) 2 +2/ 2 + 4a 2 . 



Hence 



av(# + c) 2 + y 2 = a 2 + ex. 



(2) 

Square again : 

a 2 a; 2 + 2 a 2 cx + a 2 c 2 -f- a 2 i/ 2 = a 4 + 2 a 2 ca; + c 2 » 2 , 
or 

(3) (a 2 - c> 2 + ay-= a 2 (a 2 - c 2 ). 

This is precisely the same equation that presented itself in 
the case of the ellipse ; but the locus is a curve of wholly dif- 
ferent nature. The reason is, that a and c have different 
relative values. In the ellipse, a was greater than c, and hence 
a? — c 2 was positive. It could be denoted by b 2 . Here, a is 
less than c ; a 2 — c 2 is negative, and it cannot be set equal to b 2 . 
It can, however, be set equal to — b 2 . This we will do : 

(4) a 2 -c 2 = -& 2 , or c 2 = a 2 +6 2 , 

thus defining the quantity b in the case of the hyperbola by 
the equation : 

6=Vc 2 -a 2 . 

The final equation between x and y can now be written in 
the form 



(5) 



£_£«!; 



This equation is satisfied by the coordinates of all points on 
the right-hand branch, as is seen from the way in which it 
was deduced. It is, however, also satisfied by the coordinates 
of all points on the left-hand branch. For such a point, the 



THE HYPERBOLA 127 

signs of both radicals in (1) will be reversed. Starting, now, 
with the new equation and proceeding as before, we find the 
same equation (3), which we may again write in the form (5), 
and thus the truth of the statement is established. 

Is (5) satisfied by the coordinates of still other points ? To 
answer this question, let (x, y) be any point whose coordinates 
satisfy (5). Then, starting from (5), we retrace our steps, 
admitting, each time that we extract a square root, both signs 
of the radical as conceivably possible. Thus we can be sure 
that (x, y) will satisfy one of the four equations 



± V(a + c) ; + y-± -V(x -c)-*+ y 2 = 2 a, 

corresponding to the four conceivable choices of the signs of 
the radicals : 

i) - + ; iii) - - ; 

ii) + -; iv) + +. 

If (x, y) satisfies i) or ii), the point lies on the hyperbola. 
The other two cases are impossible. For, case iii) says that a 
negative quantity is equal to a positive quantity, and case iv) 
says that F'P +FP=2 a. Now F'P + FP, being the sum of 
two sides of the triangle FPF', is greater than the third side, 
FF', or 2 c. But 2 a is actually less than 2 c. Hence we have 
a contradiction, and this case cannot arise. 

We have shown then, finally, that (5) is the equation of the 
hyperbola. 

EXERCISE 

Plot the hyperbola - ' 

25 16 
directly from its equation, taking 1 cm. as the unit of length. 

3. Axes, Eccentricity, Focal Radii. The transverse and the 
conjugate axis have already been defined in § 1. The segment 
AA' of the transverse axis is called the major aa»s,. and this 
term is also applied to its length, 2 a. The segment BB' of 
the conjugate axis, whose center is at and whose length is 





128 ANALYTIC GEOMETRY 

2 b, is called the minor axis, and this term is also applied to 

its length, 2 b. 

The major axis of an ellipse is always longer than the minor 

axis. In the case of the hyperbola, however, this is not al- 
ways true. For example, if 2 c and 2 a 
are taken as 10 and 6 respectively, then 
2 6 = 8. Thus the major axis of the 
hyperbola is to be understood as the 
principal axis, but not necessarily as the 
longer axis. 
The eccentricity of the hyperbola is defined as the number 

e = £- 
a 

Since c is greater than a, the eccentricity of a hyperbola is 
always greater than unity. 

The eccentricity characterizes the shape of the hyperbola. 
All hyperbolas having the same eccentricity are similar, differ- 
ing only in the scale to which they are drawn, and conversely ; 
cf. Exercise 8. 

The focal radii FP, F'P can be represented by simple ex- 
pressions, similar to those which presented themselves in the 
case of the ellipse. On dividing equation (2), § 2, through by 
a, we have : 

V(#+ c)' + y l = a + ex. 
Hence, when P is a point of the right-hand branch, 

(1) F'P = ex+a. 
The evaluation, 

(2) FP = ex- a, 

is obtained in a similar manner.* 

If P is a point of the left-hand branch, these formulas 
become : 

(3) F'P = -(ex + a); FP=- (ex - a). 

* P being a point of the right-hand branch, x is positive and greater 
than or equal to a ; also, e> 1. Hence ex> a, and ex — a is positive, 
as it should be. 



THE HYPERBOLA 129 

EXERCISES 

1. Find the lengths of the axes, the coordinates of the foci, 
and the value of the eccentricity for each of the following 
hyperbolas. 

(a) g- 1 = 1. Ana. 8, 6 ; (5, 0), (- 5, 0) ; 1\ 

(b) tf- y = a\ Ans. 2a, 2 a; (a V2, 0), (- a V2, 0) ; V2. 

(c) 4x--3 2 /- > = 24. (e) 5^-6^ = 8. 

(d) 2^-^ = 4. (/) 6^-9^ = 4. 

2. If the eccentricity of a hyperbola is 2 and its major axis 
is 3, what is the length of its minor axis ? Ans. 3 V3. 

3. How far apart are the foci of the hyperbola in Ex. 2 ? 

Ans. 6. 

4. What is the equation of the hyperbola whose eccentricity 
is V2 and whose foci are distant 4 from each other ? 

Ans. x l — y 2 = 2. 

5. The extremities of the minor axis of a hyperbola are in 
the points (0, ± 3) and the eccentricity is 2. Find the equa- 
tion of the hyperbola. 

6. Show that, in terms of a and b, e has the value 



_ Va* + ¥ 



e = 



7. Express b in terms of a and e. 

8. Prove that two hyperbolas which have the same eccen- 
tricity are similar, and conversely. 

9. Establish formulas (3). 

4. The Asymptotes. Two lines, called the asymptotes, stand 

in a peculiar and important relation to the hyperbola. They 

are the lines 

bx j bx 

y = — and y = 

a a 



130 



ANALYTIC GEOMETRY 




Let a point P : (x, y) move off 
y ) along a branch of the hyperbola 



(1) 



:£ _ t. — i 
a 2 62 



Fig. 7 



and let this take place, for def- 
initeness, in the first quadrant. 
The slope of the line OP is 



MP = y 
OM x 

Since the coordinates (x, y) of P satisfy (1), it follows that 

b 



(2) 

and hence 

(3) 



V 



Vcc 2 



x a * x 1 



When P recedes indefinitely, x increases without limit, and 
the right-hand side of this equation approaches the limit b/a. 
Thus we see that the slope of OP approaches that of the line OQ, 



(4) 



b 
y = -x, 

a 



as its limit, always remaining, however, less than the latter 
slope, so that P is always below OQ. 

It seems likely that P will come indefinitely near to this 
line ; but this fact does not follow from the 
foregoing, since P might approach a line 
parallel to (4) and lying below it. In that 
case, all that has been said would still be true. 

That P does, however, actually approach 
(4) can be shown by proving that the dis- 
tance PQ approaches as its limit. Now, 

PQ=MQ-MP, 
and, from (4), 

MQ = -x. 
a 




Fig. 8 



THE HYPERBOLA 131 

Furthermore, MP is the ^-coordinate of the point P on the 
hyperbola : 

a 

Hence PQ = - [x — vV — a*]. 

a 

To find the limit approached by the square bracket, we re- 
sort to an algebraic device. The value of the bracket will 
clearly not be changed if we multiply and divide it by the 
expression x + V# 2 — a 2 : 



x+Vx 2 — a 2 
But the numerator of the last expression reduces at once to a 2 

Hence 2 

x _ Va 2 - a 2 = a 



x-\-^/x —a 1 
From this form it is evident that the bracket approaches 
when x increases indefinitely; and hence the limit of PQ is 
zero, 5 * q. e. d. 

Similar reasoning, or considerations of symmetry, applied in 
the other quadrants, show that in the second and fourth 
quadrants P approaches the line 

(o) y = ~\ X > 

while in the third quadrant, as in the first, P approaches (4). 
The equations (4) and (5), of the asymptotes, can also be 
written in the form 

*_.v=o, *+'J=o. 

a b a b 



* The limit approached by the variable x — Vx 2 — a 2 can be found 
geometrically as follows. Construct a variable 
right triangle, one leg of which is fixed and of 
length a. the hypothenuse being variable and of JvP- 

length x. Then the above variable, x — Vaj 2 — a 2 , 
is equal to the difference in length between the 

hypothenuse and the variable leg. This difference obviously approaches 
as x increases indefinite! y. 




132 



ANALYTIC GEOMETRY 



It is easy to remember these equations, since they can be 
written down by replacing the right-hand side of (1) by 0, fac- 
toring the left-hand side : 



(H)(M)=°> 



and putting the individual factors equal to zero. 

The slopes of the asymptotes are b/a and — b/a. Conse- 
quently, the asymptotes make equal angles with the transverse 
axis. 

Since the ratio of b to a is unrestricted, the asymptotes can 
make any arbitrarily assigned angle with each other. If, in 
particular, b = a, this angle is a right angle, and the curve is 
called a rectangular, or equilateral, hyperbola. Its equation can 
be written in the form : 
(6) rf - f = a 2 . 

Its eccentricity is e = V2. 

Construction of the Asymptotes. Mark with heavy lines the 

major and minor axes, and through the extremities of each 

draw lines parallel to the other, 
thus obtaining a rectangle. The 
diagonals of this rectangle, pro- 
duced, are the asymptotes, since 
their slopes are clearly ± b/a. 

The diagonals of the rectangle 
have lengths equal to the distance 
2 c between the foci, for, c 2 = a 2 + b 2 

and the lengths of the sides of the rectangle are 2 a and 2 b. 

If the acute angle between an asymptote and the transverse 

axis is denoted by a, then 

e = sec a. 




Fig. 10 



EXERCISES 

1. Find the equations and slopes of the asymptotes of the 
hyperbolas of Exercise 1, § 3. Draw the hyperbolas. 



THE HYPERBOLA 133 

2. Show that the asymptotes of the hyperbola 

Ax°- - By 1 = C, 

where A, B, and C are any three positive quantities, are given 
by the equations 

VAx + y/By = 0, VAx - ^/By = 0. 

3. Find the equation of the hyperbola whose asymptotes 
make angles of 60° with the axis of x and whose vertices are 
situated at the points (1, 0), and (— 1, 0). Ans. Sx" — y 2 = 3. 

4. Show that the slopes of the asymptotes are given by 
the expression ± Ve 2 — 1. 

5. The slope of one asymptote of a hyperbola is -f. Find 
the eccentricity. Ans. e -— 1^. 

6. The distance of a focus of a certain hyperbola from the 
center is 10 cm. , and the distance of a vertex from the focus is 
2 cm. What angle do the asymptotes make with the conju- 
gate axis? Ans. 53° 8'. 

7. Show that the circle circumscribed about tne rectangle 
of the text passes through the foci. 

8. A perpendicular dropped from a focus F on an asym- 
ptote meets the latter at E. Show that OE = a, and EF = b. 

9. Find the equation of the equilateral hyperbola whose 
foci are at unit distance from the center. 

10. Find the equation of the equilateral hyperbola which 
passes through the point (—5, 4). 

5. Tangents. The method of finding the slope of an ellipse, 
Ch. IX, § 2, can be applied to the hyperbola, and it is thus 
shown that the slope of this curve, 

x 2 y 2 _H 
a*~~b>- > 
at the point (x l} y{) is 



134 



ANALYTIC GEOMETRY 



The equation of the tangent of the hyperbola at this point is 



a) 



x x x 
a'- 



62 



Theorem. TJie tangent of a hyperbola at any point bisects the 
angle betiveen the focal radii. 

To prove this proposition we recall the theorem of Plane 
Geometry which says that the bisector of an angle of a triangle 

divides the opposite side into seg- 
ments which are proportional to the 
{P:(x h y{) adjacent sides. It is easily seen 
that the converse* of this proposi- 
tion is also true, and hence it is 
sufficient for our proof to show that 

(2) ^ = Z^. 

v ; FM F'M 

We already have simple ex- 
pressions for the numerators. If 
P'- ( x i> Vi) be a point of the right-hand branch of the curve, 
then, by § 3, 

FP = efy - a ; F'F = ex x + a. 

To compute the denominators, find where the tangent at P, 
whose equation is given by (1), cuts the axis of x. Denoting 
the abscissa of M by x\ we have : 

r.2 




Fig. 



x' = 



Xl 



Now, 



FM=OF-OM=c-x', 
a* 



and 








c 


-x' 


= c- 


Xi 


x x 


Bute 


= ae, 


and 


so 




CXi 


-a* 


— a{ex x 


-a) 



Thus c — x' = — (ex x — a), 

Xi 

* Let the student prove this proposition as an exercise. 



THE HYPERBOLA 135 

and we arrive finally at the desired expression for FM : 
FM = ~{ex x - a). 
In a similar manner it is shown that 

F'M=-(ex 1 + a). 
From these evaluations it appears that 

— = - and F'P = a 

FM x 1 FM x x ' 

Hence (2) is a true equation, and the proof is complete for the 
case that P lies on the right-hand branch. Since, however, 
the curve is symmetric in the conjugate axis, the theorem is 
true for the left-hand branch also. 

Latus Rectum. The latus rectum of a hyperbola is defined 
as a chord passing through a focus and perpendicular to the 
transverse axis. The term is also applied to the length of 
such a chord. 

EXERCISES 

1. Find the slope of the hyperbola 4x 2 — ?/ 2 = 15 at the 
point (2, — 1). Ans. — 8. 

2. Find the equation of the tangent of the hyperbola of 
Ex. 1 at the point there mentioned. Ans. Sx -\- y = 15. 

3. Find the angle at which the line through the origin bi- 
secting the angle between the positive axes of coordinates cuts 
the hyperbola of Ex. 1. Ans. 30° 58'. 

4. Find the length of the latus rectum of the hyperbola 

y~=l. Ans. 4i 

16 9 

5. Find the length of the latus rectum of the hyperbola of 
Ex. 1. Ans. 15.49. 

6. Find the equation of the normal of the hyperbola 

25 144 



136 



ANALYTIC GEOMETRY 



at the extremity of the latus rectum which lies in the first 

quadrant. Ans. 25 x + 65y = 2197. 

7. Show that the length of the latus rectum of the hyperbola 



is 



2 62 



6 2 



8. Prove that the tangents at the extremities of the latera 
recta have slopes ± e. 

9. In an ellipse, the focal radii make equal angles with the 
tangent. Prove this theorem by the method employed in this 
paragraph to prove the corresponding theorem relating to the 
hyperbola. 



6. New Definition. The Directrices. The locus defined 
in Ch. VII, § 7, can now be shown to be a hyperbola when 
c > 1. The analytic treatment given there and in § 8 down 
to equation (2) and the transformation (3) holds unaltered 
for the present case. 

When, however, e > 1, the new origin, O', lies to the left of 

O, in the point ( ^— , ), and it is more natural to write 

(3) in the form 



(i) 



x' = x + 



m 



and likewise (4) as 



(2) 



s' 2 -_L = 



■ 2 -l 



€W 



2/' = 2/> 



€ , _ 1 ( £ 2 _ 1)2 

This equation passes over into the form 



(3) 

on setting 

(4) 



a? b 2 * 




a=s 



T 



6 = 



em 



Ve 2 -1 



THE HYPERBOLA 137 

Thus the locus is seen to be a hyperbola with its center, 0', 
at the point ( — o m , 0), the semi-axes being given by (4). 

The value of c is given by the equation c- = a- + b 2 . Hence 
(5) 



e 2 -l 

The eccentricity, e =c/a, is seen to be precisely e : 

e = c, 

and thus the given constant, e, turns out to be the eccentricity of the 
hyperbola. 

Finally, F is one of the foci. For, the distance from 0' to F 
is ? 

O f O+OF = -^— + m 



E 2_!^ c 2_i' 

and this, by (5), is precisely c. 

The line D is called a directrix of the hyperbola. Its dis- 
tance from the center is 

m cm 1 a 



0^ = 



■ 2 -l 



2%e Directrices. There is a second directrix, namely, the 
line D' symmetric to D in the conjugate axis. It is clear 
from the symmetry of the figure that what is true of the hy- 
perbola with respect to the focus F and the corresponding 
directrix D is equally true with respect to the focus F' and 
the directrix D'. Accordingly, the hyperbola is the locus of a 
point whose distance from a focus bears to its distance from 
the corresponding directrix a fixed ratio, the eccentricity. 

The equations of the directrices of the hyperbola, 

a 2 6 2 
are sc = - and x = — -• 



138 ANALYTIC GEOMETRY 



EXERCISES 



1. Take e = 2 and m=3, the unit of length being 1 cm. 
With ruler and compasses construct a generous number of 
points of the locus, and then draw in the locus with a clean, 
firm line.* 

2. Work out the equation of the locus of Ex. 1 directly, 
using the method of Ch. VII, § 7, but not looking at the 
formulas. Ans. 3 x 1 — y 2 + 6 x = 9. 

3. By means of a transformation to parallel axes show 
that the curve of Ex. 2 is a hyperbola whose center is at the 
point (—1, 0) and whose axes are of lengths 4 and 4V3. 

4. Show that in the general case the distances of the 
vertices, A and A', from are : 

OA = -^ % A'0= m 



€ + r €-1 

5. Collect the results of this paragraph in a syllabus, 
arranged in tabular form, giving each of the quantities, a, b, c, 
O'O, OA, A'O, OF, and F'O, in terms of m and e. 

6. Work out each of the quantities of Ex. 5 directly for 
the curve of Ex. 2 and verify the result by substituting the 
values e = 2, m = 3 in the formulas of the syllabus. 

7. Show that the tangent to the hyperbola 

16 9 

at an extremity of a latus rectum cuts the transverse axis in 
the same point in which this axis is cut by a directrix. 
8. The same for any hyperbola. 

* The footnote of p. 114 applies in the present case with the obvious 
modification that the distance of the parallel from D must now be half 
the radius of the circle. Moreover, two parallels to D must now be drawn, 
the second one, as soon as the radius has increased sufficiently, giving 
points on the left-hand branch. 



THE HYPERBOLA 139 

9. Prove directly that, if P is any point of the hyperbola 

the ratio of its distance from a focus to its distance from the 
corresponding directrix equals the eccentricity. 

10. Prove that the ratio of the distance between the foci of 
a hyperbola to the distance between the directrices equals the 
square of the eccentricity. 

7. The Parabola as the Limit of Hyperbolas. Summary. 

Equation (3) of Ch. VII, § 7, namely, 

(1) (1 — e 2 )x- + y 1 — 2mx + m- = 0, 

represents a hyperbola when e > 1 and a parabola when e = 1. 
If, then, we let c approach 1 through values greater than 1, 
the hyperbola which (1) represents will approach a parabola 
as its limiting position. 

Suppose, for example, that we take m = 2 and let e take on 
successively the values 2, 1-J-, 1-J-, 1-J-, •••. Drawing the corre- 
sponding hyperbolas, we find that, whereas the directrix D 
and the right-hand focus F are always fixed, the center and the 
left-hand vertex keep receding to the left, and that their 
distances from 0, namely, 

, = ^n_ A , Q== m 



-1' 6-1' 

increase without limit. Thus, when e approaches 1, the left- 
hand branch of the hyperbola recedes indefinitely to the left 
and disappears in the limit, whereas, meanwhile, the right- 
hand branch gradually changes shape and in the limit becomes 
the parabola whose directrix is D and whose focus is F. 

Summary. Let us now combine the results of § 6 with those 
of § 8, Ch. VII. We have proved that equation (1) repre- 
sents an ellipse, a parabola, or a hyperbola, according as e< 1, 
e = 1, or c > 1. In case of the ellipse and the hyperbola the 



140 ANALYTIC GEOMETRY 

constant c turned out to be the eccentricity e. We are led 
then to give to the parabola an eccentric! ty ; namely, 
e = e = l. 

Theorem. Tlie locus of a point ivhich moves so that its dis- 
tance from a fixed point bears to its distance from a fixed line, 
not passing through the fixed point, a given ratio e is an ellipse, 
a parabola, or a hyperbola, according as e is less than, equal to, 
or greater than unity. In every case the constant e equals the 
eccentricity. 

Since always e = e, we may suppress e in future work, and 
use e exclusively. Thus equation (1) becomes 

(2) (1 - e*)x ' + f' - 2 mx + m 2 = 0. 

The theorem furnishes a blanket definition for the ellipse, 
parabola, and hyperbola, which might have been used instead 
of the separate definitions which we have given. It should 
be noted, however, that this blanket definition does not include 
the circle. For, if we set e = in (2), the equation reduces to 

(x-my+y2=0, 

which represents merely the focus F : (m, 0). 

The fact that the blanket definition does not yield a circle 
as a special case in no way discredits the circle as the limiting 
form of an ellipse when the eccentricity approaches zero, 
Ch. VII, § 1. The reason that a circle cannot be defined in 
the new manner is because it has no directrices. When the 
eccentricity of an ellipse approaches zero, the major axis 
remaining constant, the distance a/e of the directrices from 
the center increases indefinitely, so that in the limit, when the 
ellipse becomes a circle, the directrices have disappeared.* 

* It is, of course, possible to obtain the circle as a limiting curve ap- 
proached by ellipses defined in the new way. If the points F and A of 
Fig. 12, Ch. VII, are held fast and m is allowed to increase indefinitely, 
then it can be shown that e approaches zero and that a and b both approach 
the fixed distance AF. Thus the variable ellipse approaches a circle as 
its limit. 



THE HYPERBOLA 



141 



8. Hyperbolas with Foci on the Axis of y. Conjugate 
Hyperbolas. Let the student show that the equation of the 
hyperbola whose foci are at the points (0, ± C) on the axis 
of y and the difference of whose focal radii is 2 B is 



E- — UL 
A* B* 



where 



-1, 

C 2 = A> + B\ 



The transverse axis of this hyperbola is the axis of y ; the 
conjugate axis, the axis of x. The length of the major axis is 
2 B ; that of the minor axis, 2 A. The eccentricity is C/B and 
the asymptotes have the equations, 



£-£ = 

A B 



and 



£ + 1 = 0. 
A^B 



Conjugate Hyperbolas. The two hyperbolas, 



a? 2/ 2 



-2-=bl 



and 



a 1 b- 



have the same asymptotes. The transverse axis of each is the 
conjugate axis of the other, and the major axis of each is the 
minor axis of the other. 

Taken together, the two 
hyperbolas form what is 
called a pair of conjugate 
hyperbolas. The relation- 
ship between them is per- 
fect in its duality. We 
say, then, that each is the 
conjugate of the other. 

The two hyperbolas to- 
gether are tangent exter- 
nally at their vertices to the rectangle of § 4 at the mid-points 
of its sides. Moreover, all straight lines through the common 
center O, except two, meet one hyperbola or the other in two 
points, and the segment thus terminated is bisected at 0. 




Fig. 13 



142 ANALYTIC GEOMETRY 

The student should compare these facts with the correspond- 
ing ones concerning a single ellipse and the circumscribed 
rectangle. 

EXERCISES 

1. Find the coordinates of the foci, the lengths of the axes, 
the slopes of the asymptotes, and the value of the eccentricity 
for each of the hyperbolas : 

(tt) f~S = ~ 1; (C) 2/ 2 -^ = 4 ' 

(b) 5a 2 -4^+20=0; (d) Sx^ - 2y* + 6 = 0. 

Draw an accurate figure in each case. 

2. What are the equations of the hyperbolas conjugate to 
the hyperbolas of Ex. 1 ? 

3. Find the equation of the hyperbola whose vertices are in 
the points (0, ± 4) and whose eccentricity is -§-. 

Ans. 4^-5^4-80 = 0. 

4. Find the equation of the hyperbola the extremities of 
whose minor axis are in the points ( ± 3, 0) and whose eccen- 
tricity is J . 

5. Prove that the sum of the squares of the reciprocals of 
the eccentricities of the two conjugate hyperbolas 

9 16 ' 9 16 

is equal to unity. 

6. Prove the theorem of Ex. 5 for the general pair of conju- 
gate hyperbolas. 

7. Show that the foci of a pair of conjugate hyperbolas 
lie on a circle. 

9. Parametric Representation, It is possible to construct a 
hyperbola, given its axes, AA' and BB', by a method much 
like that of Ch. VII, § 10, for the ellipse. 



iHE HYPERBOLA 



143 



\P:ix,V) 



Let the two circles, C and C", and the ray from 0, be drawn 
as before. At the point L 
draw the tangent to C", and 
mark the point Q where the 
ray cuts this line. At R draw 
the tangent to C and mark the 
point S where this tangent 
cuts the axis of x. 

The locus of the point 
P : (x, y), in which the paral- 
lel to the axis of x through 
Q and the parallel to the 
axis of y through S intersect, is the hyperbola. 




For, 






OR = a, 


OL = b, 


and 


x = 


OS 


= a sec cf>, 


y = LQ = b tan <£ 


Hence 






x 

- = secd>, 

a 


^ = tan <£, 
b 


and since 






sec 2 <f> — 


tan 2 <f> — 1, 


it follows that 




tf_ 


£=i. 



a 2 &-' 

Conversely, any point (x, y) whose coordinates satisfy this 
equation is seen to lead to an angle <£, for which the above 
formulas hold. 

We thus obtain the following parametric representation of 
the hyperbola : 

x = a sec <f>, y = b tan <f>. 

The circle C, constructed on the major axis of the hyperbola 
as a diameter, is known as the auxiliary circle of the hyperbola, 
and the angle <£ is called the eccentric angle. 



EXERCISES 
1. Carry out the construction described above for the cases 
(a) a = 3 cm., 6 = 2 cm. 



144 



ANALYTIC GEOMETRY 



(6) a = 3cm., b = 3 cm. 
(c) a = 2 cm., 6 = 3 cm. 
2. Obtain a parametric representation of the hyperbola 



x2 _ t_ _ _ i 



^4 2 £ 2 

10. Conic Sections. The ellipse (inclusive of the circle), the 
hyperbola, and the parabola are often called conic sections, 

because they are the curves 
in which a cone of revolution 
is cut by planes. 

Suppose a plane M cuts only 
one nappe of the cone, as is 
shown in the accompanying 
drawing. Let a small sphere 
be placed in the cone near 0, 
tangent to this nappe along a 
circle. It will not be large 
enough to reach to the plane 
M. Now let the sphere grow, 
always remaining tangent to 
the cone along a circle. It 
will finally just reach the 
plane. Mark the point of 
tangency, F, of the plane M 
with the sphere, and also the 
circle of contact, C, of the 
sphere with the cone. 
As the sphere grows still larger, it cuts the plane M, but 
finally passes beyond on the other side. In its last position, in 
which it still meets M } it will be tangent to M. Let the point 
of tangency be denoted by F', and the circle of contact of the 
sphere with the cone by C. 

Through an arbitrary point P of the curve of intersection of 
M with the cone passes a generator OP of the cone ; let it cut 
C in R and C in R'. Then RR'> being the slant height of 




Fig. 15 



THE HYPERBOLA 145 

the frustum * cut from the cone by the planes of C and C-, is 
of the same length, 2 a, for all points P. 

Join P with F. Then PF and PR, being tangents from P 
to the same sphere, are equal. Similarly, PF' and PR' are 
equal. Hence 

FP + F'P =RP+ R'P= RR', 
or FP+F'P = 2a. 

But this locus is by definition an ellipse with its foci 
at F and F', and hence the proposition is proved for the case 
that M cuts only one nappe, the intersection being a closed 
curve. 

If the plane M cuts both nappes, but does not pass through 
0, it is a little harder to draw the figure, one sphere being 
inscribed in the one nappe, the other, in the other nappe. 
A similar study shows that here the difference between 
FP arid F'P is equal to RR f , and hence the locus is a 
hyperbola. 

The parabola corresponds to the case that M meets only one 
nappe, but does not cut it in a closed curve. This case is 
realized when M does not pass through and is parallel to a 
generator of the cone. 

Let L be a line which is perpendicular to the axis of the 
cone in a point of the axis distinct from the vertex. As a 
plane, M, rotates about L, it will cut from the cone all three 
kinds of conies. This will still be true if we take, as L, any 
line of space which does not pass through the vertex and is 
not parallel to a generator. 

11. Confocal Conies. Two conies are said to be confocal if 
they have the same foci ; in the case of two parabolas, we de- 
mand, further, that they have the same axis. 

* No technical knowledge of Solid Geometry beyond the definitions of 
the terras used (which can be found in any dictionary) is here needed. 
On visualizing the figure, the truth of the statements regarding the space 
relations becomes evident. 




146 ANALYTIC GEOMETRY 

Consider an ellipse and a hyperbola which are confocal. 
They evidently intersect in four points.* 

Let P be one of these points. Join P with F and F'. 
Then FP and F'P are focal radii both of the ellipse and of the 
hyperbola. Now, the tangent to a hyper- 
bola at any point not a vertex bisects the 
angle between the focal radii drawn to 
that point, § 5 ; and the normal to an 
ellipse at any point not on the transverse 
axis bisects the angle between the focal 
radii drawn to that point, Ch. VII, § 4. 
It follows, then, that the tangent to 
the hyperbola at P and the normal to the ellipse at this 
point coincide. Hence the two curves intersect at right 
angles, or orthogonally, as we say. We have thus proved the 
following 

Theorem. A pair of confocal conies, one of which is an el- 
lipse and the other a hyperbola, cut each other orthogonally. 

Confocal Parabolas. Consider two parabolas having the 
same focus and the same axis. If both open out in the same 
direction, they have no point in common. If, however, they 
open out in opposite directions, they intersect in 
two points which are symmetrically situated with 
respect to the axis. 

In the latter case, the parabolas intersect orthogo- 
nally, as has already been proved analytically; cf. 
Ch. VI, § 3, Ex. 10. 

This result could have been forecast, as a conse- fig. 17 
quence of the relations established in § 7. For, if 
one focus, F, and the two corresponding directrices of a pair 
of confocal conies, consisting of an ellipse and a hyperbola, 
are held fast, and if the other focus is made to recede in- 
definitely, each of the conies approaches a parabola. But the 

* Let the student satisfy himself that two confocal ellipses do not in- 
tersect, and that the same is true of two confocal hyperbolas. 




THE HYPERBOLA 



147 




Fig. 18 



conies always intersect orthogonally, and so the same will be 
true of the limiting curves, the parabolas. 

To obtain a prescribed pair of parabolas, like those described 
above, as limiting curves, it is necessary merely to choose the 
two confocal conies so that the directrices corresponding to F 
are at the proper distances from F. 

Mechanical Constructions. It is possible to draw with ease a 

large number of confocal ellipses by the method set forth in Ch. 

VII, § 1. Let thumb tacks be inserted at F 

and F', but not pushed clear down. Let a 

thread- be tied to the tack at F, passed round 

the tack at F', and held fast at M. Then an 

ellipse can be drawn with F and F ' as foci. 
Now let the thread be unwound at F' 

and drawn in or paid out slightly, so that 

the length of the free thread between F and F' is changed. 

On repeating the above construction, a second ellipse with 

its foci at F and F' is obtained ; and so on. 

There is an analogous construction for a hyperbola, which 

has not yet been mentioned. Tie a thread to a pencil point,* 
pass the thread round the pegs at F and 
F' as shown, hold the free ends firmly 
together at M, and, keeping the thread 
taut by pressing on the pencil, allow M to 
move. The pencil then obviously traces 
out a hyperbola. 

By pulling one end of the 
thread in slightly at M, or by 

paying it out, and then repeating the construction, 

a new hyperbola with the same foci is obtained ; 

and so on. 

Parabolas. The accompanying figure suggests 

a means for drawing a parabola mechanically. 

* To keep the thread from slipping off, cut a groove in the lead, such 

as would be obtained if the pencil were turned about its axis in a lathe 

and the point of a chisel were held against the lead close to the wood. 





Fig. 20 



148 ANALYTIC GEOMETRY 

A ruler, D, is held fast and a triangle, T, is allowed to slide 
along the ruler. A thread is tied at F and Q, and a pencil 
point, P, keeps the thread taut and pressed against the 
triangle. 

EXERCISES 

1. Show that the conies, 

^- + ^=1 and £-£-1, 
24^8 4 12 ' 

are confocal. 

2. Prove that the equation, 



9+A " 5 + A ' 

represents an ellipse for eacn value of A greater than — 5 and 
represents a hyperbola for each value of A between — 9 and 
— 5. Show that all these ellipses and hyperbolas are confocal, 
with the points (± 2, 0) as foci. 

3. For what values of A. does the equation 



a 2 + A b 2 + A 



= 1, 



where a and 6 are given positive constants such that a > b, 
represent i) ellipses ? ii) hyperbolas ? Show that all these 
conies are confocal. 

4. Draw a set of confocal ellipses and hyperbolas. 

5. Draw a set of confocal parabolas, all having the same 
transverse axis, some opening in one direction, some in the 
other. 

EXERCISES ON CHAPTER VIII 

1. The axes of a hyperbola which goes through the points 
(1, 4), (— 2, 7) are the axes of coordinates. Find the equation 
of the hyperbola. Ans. y 2 — 11a; 2 = 5 

2. Show that the hyperbolas defined by the equation 

4:X 2 -5y 2 = c, 



THE HYPERBOLA 149 

where c is an arbitrary constant, not zero, all have the same 
asymptotes. 

3. How many hyperbolas are there with the lines 

3a? 2 -16^ = 

as asymptotes ? Find an equation which represents them all. 

Ans. 3 x 1 — 16 y 1 = c, c =£ 0. 

4. What is the equation of all the rectangular hyperbolas 
with the axes of coordinates as axes ? 

5. A hyperbola with the lines 4^ — ^ = as asymptotes 
goes through the point (1, 1). What is its equation ? 

Ans. 4 x — y" 1 = 3. 

6. The asymptotes of a hyperbola go through the origin 
and have slopes ± 2. The hyperbola goes through the point 
(1, 3). Find its equation. Ans. 4 a;" 5 — y 2 = — 5. 

7. The two hyperbolas of Exs. 5 and 6 have the same 
asymptotes, but lie in the opposite pairs of regions into which 
the plane is divided by the asymptotes. Show that the sum 
of the squares of the reciprocals of their eccentricities equals 
unity. 

8. Prove that of the hyperbolas of Ex. 2 those for which 
c is positive are all similar, and that this is true also of those 
for which c is negative. If e is the common value of the ec- 
centricity of the hyperbolas of the first set and e f is that of 
the hyperbolas of the second set, show that 

(1) I + A=l. 

v } e 2 e' 2 

9. Prove that the relation (1) is valid for the eccentricities 
of any two hyperbolas which have the same asymptotes but 
lie in the opposite regions between the asymptotes. 

10. Show that two hyperbolas which are related as those 
described in the previous exercise have the same eccentricity 
if and only if they are rectangular hyperbolas. 



150 ANALYTIC GEOMETRY 

11. A hyperbola with its center in the origin has the eccen* 
tricity 2. Find the equations of the asymptotes, (a) if the 
foci lie on the axis of x ; (6) if the foci lie on the axis of y. 

Am. (a) 3^-^ = 0; (b) a- -3^ = 0. 

12. What is the equation representing all the hyperbolas 
which have their centers in the origin and eccentricity 2, 
(a) if the foci lie on the axis of x ? (6) if the foci lie on the 
axis of y ? Show that in either case the vertices lie midway 
between the center and the foci. 

13. Prove that the vertices of the hyperbola 

a 2 b* 

subtend a right angle at each of the points (0, ± b) when and 
only when the hyperbola is rectangular. What is the corre- 
sponding theorem in the case of the ellipse? 

14. The projections of a point P of a hyperbola on the 
transverse and conjugate axes are P 2 and P 2 . The tangent at 
P meets these axes in 2\ and T 2 . Show that OP 1 • OT x = a 2 
and OP 2 - OT 2 = — b\ where is the center of the hyperbola 
and a and b are the semi-axes. 

15. Prove that the segment of a tangent to a hyperbola be- 
tween the point of contact and a directrix subtends a right 
angle at the corresponding focus. 

16. The projection of a point P of a hyperbola on the 
transverse axis is P x and the normal at P meets this axis at 
Ni. Show that the ratio of the distances of the center from 
iVi and P x equals the square of the eccentricity^ 

17. Prove that the line joining a point P of a hyperbola 
with the center and the line through a focus perpendicular to 
the tangent at P meet on a directrix. 

18. Find the equation of the circle which is tangent to a 
hyperbola at the upper ends of the two latera recta. 

19. Let be the center, A a vertex, and F the adjacent 
focus of a hyperbola. The tangent at a point P meets the 



THE HYPERBOLA 151 

transverse axis at T and the tangent at A meets OP at V. 
Show that TV is parallel to AP. 

20. Show that an asymptote, a directrix, and the line through 
the corresponding focus perpendicular to the asymptote go 
through a point. 

21. A line through a focus F parallel to an asymptote meets 
the hyperbola at P. Show that the tangent at P, the other 
asymptote, and the line of the latus rectum through F meet in 
a point. 

22. Let F be a focus and D the corresponding directrix of 
a hyperbola. A line through a point P of the hyperbola parallel 
to an asymptote meets 1) in the point K. Prove that the tri- 
angle FPK is isosceles. 

Exercises 23-33. In proving the theorems in these exercises 
it will, in general, be necessary to make actual use of the 
equation expressing the fact that a certain point lies on the 
hyperbola. 

23. The tangent to a hyperbola at a point P meets the tan- 
gent at one vertex in Q. Prove that the line joining the other 
vertex to P is parallel to the line joining the center to Q, 

24. Let F be a focus and D the corresponding directrix of a 
hyperbola. Prove that the segment cut from D by the lines 
joining the vertices with an arbitrary point on the hyperbola 
subtends a right angle at F. 

25. Prove that the product of the distances of the foci of a 
hyperbola from a tangent is constant, i.e. independent of the 
choice of the tangent. 

26. Let A and A' be the vertices of a rectangular hyperbola 
and let P and P' be two points of the hyperbola symmetric in 
the transverse axis. Prove that AP is perpendicular to A'P' 
and that AP' is perpendicular to A'P. 

27. Show that the product of the focal radii to a point on a 
rectangular hyperbola is equal to the square of the distance of 
the point from the center. 



152 ANALYTIC GEOMETRY 

28. Prove that the angles subtended at the vertices of a 
rectangular hyperbola by a chord parallel to the conjugate axis 
are supplementary. 

29. Prove that the product of the distances of an arbitrary 
point on a hyperbola from the asymptotes is constant, i.e. 
the same for every choice of the point. 

30. A line through an arbitrary point P on a hyperbola 
parallel to the conjugate axis meets the asymptotes in M and 
JV. Show that the product of the segments in which P divides 
MNis constant. 

31. Prove that the segment of a tangent to a hyperbola cut 
out by the asymptotes is bisected by the point of contact of 
the tangent. 

32. Show that the tangent to a hyperbola at an arbitrary point 
forms with the asymptotes a triangle which has a constant area. 

33. The tangent to a hyperbola at a point P meets the tan- 
gents at the vertices in ifcf and AT. Prove that the circle on 
MN as a diameter passes through the foci. 

Loci 

34. Find the locus of a point whose distance from a given 
circle always equals its distance from a given point without 
the circle. First give a geometric construction, with ruler and 
compass, for points on the locus. Then find the equation of 
the locus. 

35. The base of a triangle is fixed and the product of the 
tangents of the base angles is a negative constant. What is 
the locus of the vertex ? 

36. A line moves so that the area of the triangle which it 
forms with two given perpendicular lines is constant. Find 
the locus of the mid-point of the segment cut from it by these 
lines. 

Ans. Two conjugate rectangular hyperbolas, with the given 
lines as asymptotes. 



THE HYPERBOLA 153 

37. Given a fixed line L and a fixed point A, not on L. A 
point P moves so that its distance from L always equals the 
distance AQ, where Q is the foot of the perpendicular dropped 
from P on L. What is the locus of P? 

38. What is the locus of the point P of the preceding exer- 
cise, if the ratio of its distance from L to the distance AQ is 
constant ? 



CHAPTER IX 

CERTAIN GENERAL METHODS 

1. Tangents. Let it be required to find the tangent line to 
a given curve at an arbitrary point. 

In the case of the circle the tangent is perpendicular to the 
radius drawn to the point of tangency. But this solution is 
of so special a nature that it suggests no general method of 
attack. A general method must be 
based on a general property of tan- 
gents, irrespective of the special curve 
considered. Such a method is the 
following. Let P be an arbitrary 
point of a given curve, C, at which it 
is desired to draw the tangent, T. Let 
a second point, P', be chosen on 0, 
and draw the secant, PP'. As P' 
moves along O and approaches the 
fixed point P as its limit, the secant rotates about P as a pivot 
and approaches the tangent, T, as its limiting position. Thus 
the tangent appears as the limit of the secant. 

If, now, in a given case we can find an expression for the 
slope of the secant, the limit approached by this expression 
will give us the slope of the tangent. The slope of the tangent 
to the curve at P we shall call, for the sake of brevity, the 
slope of the curve at P. 

Example 1. Find the slope of the curve 

(1) y = x* 

at a given point, P. 

154 




CERTAIN GENERAL METHODS 



155 



Let the coordinates of P be 
(*i, Vi) 5 those of P , (x', ?/'), 
or (x\ + h, yi + k). Then 

PQ = h, QP = fc, 

and we have, for the slope of 
the secant PP'. the expression : 

(2) W=|, 




where r' 
then, 

(3) 



£ QPP'. The slope of the tangent line, T, at P is, 

k 



tan t = lini tan r' = lini 
p'±p /i=o 7i 



where t= "%. QPT. The sign = is used to mean "approaches 

us its limit." and the expression : lim tan r'. is read : " the 

p'=p 

limit of tan/, as P' approaches P." 

Suppose, for example, that P is the" point (1, 1). Let us 
compute k and tan r' for a few values of h. Here, x x = l and 
y x = 1. If h = .1, then 

a;' = a?! + ft = l-.l, 
^ = y 1 +V=(ii)»=L21 > 
& = .21, 

.1 



and hence 



tan 



2.1. 



Next, let P' be the point for which 
a;' = 1.01. 
Then y' = 1.0201, 

h = .01, k = .0201, 



and hence 



, , .0201 OAi 
tan t' = — - — = 2.01. 



Let the student work out one more case, taking x' = 1.001. 
He will find that here k = .002001 and 



tanr' = 2.001. 



15b 



ANALYTIC GEOMETRY 



These results can be presented conveniently in the form of 
a table : 



* 


k 
t 


tanr' = £ 
h 


.1 


.21 


2.1 


.01 


.0201 


2.01 


.001 


.002001 


2.001 



The numbers in the last column appear to be approaching 
nearer and nearer to the limit 2 ; in other words, the slope of 
the curve in the point (1, 1) appears to be 2. Let us prove 
that this is actually the case. Since the proof is just as simple 
for an arbitrary point P, we will return to the general case. 

The point P being a point of the curve (1), its coordinates 
(x lf yj must satisfy that equation. Hence 

(4) y x = x x \ 

Similarly, for the point P' whose coordinates are (a?i + /i, 
ft + *) : 

or 

(5) y x + k = a?! 2 + 2 xji + h*. 

Subtracting (4) from (5), we get : 

k=2x l h + h 2 . 



Consequently, 



tanT' = -=2a; 1 + fc. 
h 



Now let P' approach P; h will then approach 0, and we 
shall have 



But 



lim tan t' = lim - = lim (2 x l -\-h). 

p-=P fc=0 h h±Q 



lim tan t' = tan t, and lim (2xi + h)=2 x x . 

P'±P fcM) 



Hence 



tanr == 2aji. 



CERTAIN GENERAL METHODS 157 

We can say, then, that the slope of the curve (1), at an arbi- 
trary point P : (x u yi) on it, is 

\ = 2x v 

If, in particular, P is the point (1, 1), the slope of the tan- 
gent there is \ = 2 • 1 = 2, and thus the indication given by 
the above table is seen to be borne out. 

Example 2. Find the slope of the curve 

a 2 
(6) ,-i 

at an arbitrary point P : (x l9 y^) of the" curve. 

Denote, as before, the coordinates of a second point, P', by 

<xf = xi + h, y' = y x + k. 

Then, since P and P' lie on the curve, 

a 2 
Vi = - 

and yi + k = 



Hence fc == 



#! + & 

a 2 a 2 



a?x + h Xi 

Nothing is more natural than to reduce the right-hand side 
of this equation to a common denominator. Thus 

k= - alt 

x 1 (x 1 + h) 

Consequently, 

for, / * -<* 2 

tan t = - = 



& «!(«! -f- h) 
We are now ready to let P' approach P: 



lim tan t' = lim 



P'=P *=0 Xi(x x + ft) 

The limit approached by the right-hand side is obviously 

— a 2 /x l 2 , and so 



tanr = - — « 
x x 2 



158 ANALYTIC GEOMETRY 

We have, then, as the final result : The slope of the curve (6), 
at an arbitary point (x h y{) on it, is 

x— «*. 

Equation of the Tangent. Since the tangent to the curve (1), 
y = x*, 

at the point (1, 1) has the slope 2, its equation is 

y — 1 = 2 (x - 1), or 2 x — y — 1 = 0. 

Similarly, the equation of the tangent to the curve (1) at an 
arbitrary point P : (x l9 y x ) is 

y-y A = 2x 1 (x-x 1 ), 
or 

y-y 1 = 2x 1 x-2x 1 2 . 

This equation may be simplified by use of the equality, 



Vi = *h 2 , 



which says that the point P lies on the curve. For, if we re- 
place the term 2# 1 2 by its equal, 2y lf and then combine the 
terms in y 1} the equation becomes 

y 4-2/i = 2 x x x. 



This equation of the tangent is of the first degree in x and y, 
as it should be. The quantities x 1 and y x are the arbitrary, 
but in any given case fixed, coordinates of P and are not 
variables. 

Equation of the Normal. The line through a point P of a 
curve perpendicular to the tangent at P is known as the 
normal to the curve at P. 

Since the tangent to the curve y = x 2 at the point (1, 1) 
has the slope 2, the normal at this point has the slope — i 
Consequently, the equation of the normal is 

*/-l=- !(>-!), or b + 2#-3 = 0. 



CERTAIN GENERAL METHODS 159 



EXERCISES 

1. Determine the slope of the curve y = x* — x at the point 
(3, 6). First make out a table like that under Example 1, and 
hence infer the probable slope. Then take an arbitrary point 
(#i> 2/i) on the curve and determine the actual slope at this 
point by finding 

lim— • 
a=bo h 

2. The same for the curve Sy = Sx 3 at the point (2, 3). 

3. The same for the curve y = 2x 2 — 3x + l at the point 
(1, 0). 

Find the slope of each of the following curves at an arbitrary 
point P : (x 1} y^. No preliminary study of a numerical case, 
like that which gave rise to the table under Example 1, is 
here required. 



4. 


y = x* 1 — 3 x 


+ 1. 








Ans. A = 2 #! — 3* 


5. 


y = 2x -x 


-4. 




7. 


y 


= 4^ -2a; 2 + 5. 


6. 


y — a? — x. 






8. 


y 


= jb 3 -f a,*- -f x + 1. 


9. 


y = x 3 +px-\- q. 








^Ins. A = 3a; 1 2 -fp. 


10. 


y — x^ — o? 1 . 










^Ins. A = 4 a?! 3 . 


11. 


_ 1 
V X 2 ' 










.4ws. A = - 

a?! 3 


12. 


a 4 

X* 












13. 


1 










-4tts. A = — -. 

(l-*i) 2 


14. 


3 a -4 










^ ns \ _ 8 




(3^i-4)2 


15. 


y = ax" -\- bx -f- c. 








^Ins. A = 2 ax x + b. 


16. 


y = ax 3 + bx 1 4- ex 


+ d. 








17. 


y = x n , ()l, a 


positive in 


iteger) 




Aiis. A = nxf'K 


18. 


y = cx n . 













160 ANALYTIC GEOMETRY 

Find the equations of the tangents to the following curves 
at the points specified. In each case reduce the equation ob- 
tained to the simplest form. 

19. The curve of Ex. 1 at the points (3, 6) ; (x h y{). 

Ans. 5x — y — 9 = 0; (2x x — l)x — y — x 1 2 = 

20. The curve of Ex. 3 at the points (x 1} y t ) ; (1, 0). 

21. The curve of Ex. 4 at the points (ie h y±) ; (— 1, 5) 

22. The curve of Ex. 11 at the points (1, 1) ; (x l} y x ) 

23. The curve of Ex. 17 at the point (x 1} y^. 

Ans. nxi~ l x — y — (n — l)y v = 0. 

24. The curve of Ex. 13 at the point whose abscissa is 2. 

25. • The curve of Ex. 14 at the point whose abscissa is 4. 

26. Find the equations of the normals to the curves of Exs. 
21, 22 at the designated points. 

2. Continuation. Implicit Equations. We have applied the 
general method to curves whose equations are given in the 
form : y = a simple expression in x. More precisely, this 
" simple expression " has each time been a polynomial (or even 
a monomial), or the ratio of two such expressions. 

But even the simplest forms of the equations of the conies 
are, as a rule, such that, if the equation be solved for y, radi- 
cals will appear. In such cases, the following method of treat- 
ment can be used with advantage. 

The Parabola. Let it be required to find the slope of the 
parabola 

(1) y* = 2mx 

at any point P : (x 1} y^) on the curve. 

We will treat first a numerical case, setting m = 2 : 

(2) y2 = 4 x. 
Since P is on the curve, we have 

(3) 2/i 2 = 4a 1 . 



CERTAIN GENERAL METHODS 161 

Since P' : (x x + h, y x + k) is also on the curve, we have : 

or 

(4) y{- + 2 Vl k + W< = 4^ -b 4ft. 

Subtract (3) from (4) : 

2^ + ^ = 4k 

Divide this equation through by h, to obtain an equation for 
tan t' = k/h : 

2y 1 -+k-=±, or 2 ^ tan r' + fc tan t' = 4. 

Solve the latter equation for tan r' : 

4 



tan t' = 



2 Vl + k 

We are now ready to let P' approach P as its limit. This 
means that h and k both approach 0. We have, then, 

lim tan t' = lim -, 

p-=p a=o 2y 1 + k 

or 

tan T = -^ = i. 

It has been tacitly assumed that y 1 ^= 0. If y x = 0, then 
tanr' increases indefinitely as h, and with it k, approaches 
zero. Thus the tangent line is seen to be perpendicular to the 
axis of x at this point, as obviously is, in fact, the case, since 
the point is the vertex of the parabola. 

The student will now carry through by himself the corre- 
sponding solution in the general case of equation (1). He 
will arrive at the result : The slope \ of the parabola 

y l — 2mx 

at an arbitrary point (x^, y x ) of the curve is 

(5) X = ™. 



162 ANALYTIC GEOMETRY 

The Ellipse. The treatment in the case of the ellipse, 

is precisely similar. Writing (6), for convenience, in the 
form 

(7) frx+ ay^ = ab'\ 

we are led to the following equations * : 

(8) b% 2 + a 2 y x 2 = a 2 5 2 ; 
b 2 (x L + h) 2 -\- a 2 ^ + A;) 2 = a 2 b 2 ,' 

or 

(9) bW + a 2 y x 2 + 2 fc^ft + 2 a 2 ^ + 6 2 /* 2 +a 2 fc 2 = a 2 6 2 . 

Subtract (8) from (9) : 

2 &V1 + 2 a^fc + & 2 /* 2 + a 2 k 2 = 0. 
Divide by h : 

2 fc 2 ^ + 2 « 2 2/!- + 6 2 /i + a 2 Jc - = 0, 
h h 

or 2 62^ + 2 a 2 2/i tan t' + bVi + a 2 & tan r' = 0. 

Solve this equation for tan r' : 

tant >__2M5 L± ft». 

2 a 2 2/x + a 2 & 

Now let P' approach P as its limit : 

lim tan t' = lim — • 

p=p h±v 2a 2 y 1 + a 2 ft 

„ . 2b*Xi b 2 x L 

Hence tan t = — — — - = — - • 

2a 2 Vl a 2 y x 

We have thus obtained the result : The slope \ of the ellipse 

t + t^l 
a 2 b 2 

at an arbitrary one of its points (x x , y x ) is 

(10) x = -.^« 

* The student will do well to paraphrase the text at this point with a 
numerical case, — say, 4 x 2 + 9 y 2 — 36. 



CERTAIN GENERAL METHODS 163 

The Hyperbola. The treatment is left to the student. The 
result is as follows. 

Tlie slope X of the hyperbola 

a 2 5 2 ~ 
at an arbitrary one of its points (o&u y^) is 

(11) X=^i. 

Equation of the Tangent. Since the slope of the ellipse at 
the point (a? l5 y x ) is — b 2 xja 2 yi, the equation of the tangent at 
Oi, yi) is 

or, after clearing of fractions and rearranging terms, 

b 2 x^x + ah/fl = ft 2 ^ 2 + a 2 ?/i 2 . 

If we divide both sides of this equation by a 2 6 2 , we have 

afrp | .ViV = a?i 2 , .Vi 2 . 
a 2 6 2 a 2 6 2 ' 

But, since the point (a^, y x ) lies on the ellipse, it follows that 

a 2 6 2 
and the equation of the tangent becomes 

*& , y&--\ 

a 2 b 2 
The equation of the tangent to the ellipse 

^ + ^ = 1 
a 2 b 2 
at the point (x lt yi) is 

(12) ' M + i*«l 
v ; a 2 6 2 

In a similar manner let the student establish the equations 
of the tangents to the hyperbola and the parabola. 



164 ANALYTIC GEOMETRY 

The tangent to the hyperbola 





X 2 

a 2 


6 2 


at the point (x h 


, 2/1) has the equation 


(13) 


x t x 
~oJ 


6 2 


The tangent i 


lo the parabola 






y 2 


= 2mx 



at the point (x 1: 2/1) has the equation 
(14) y$ = m(x + x x ). 

EXERCISES 

Find the slope of each of the following six curves at an 
arbitrary one of its points, applying each time the method 
set forth in the text. 

1. 2a 2 4.3?/2 = 12. 3. i/ 2 = 12 a. 

2. x 2 — 42/ 2 = 4. 4. x 2 — 2/ 2 = a\ 

5. Ax 2 4- By 2 — C, where A, B, C are all positive. 

6. y 2 = Ax + B, where A 4^ 0. 

7. Find the slope of the parabola y 2 + 2 y = 6 a; at the point 

3 
(«i, 2/i)- -4ris. A. = — — . 

2/1 + 1 

8. What is the slope of the parabola of Ex. 7 at the origin? 

Ans. 3. 

9. Find the slope of the curve 

x*-y 2 -3x + 4y = 
at the origin. Ans. X= f . 

Suggestion. First find the slope at an arbitrary point (x ly 2/1). 
Then substitute in the result the coordinates of the origin. 

10. What angle does the curve 

2x*-3y* + x-y + 1 = 
make with a parallel to the axis of x at the point (1, 1) ? 

Ans. 45°. 



CERTAIN GENERAL METHODS 



105 



11. Find the slope of the curve xy = a 2 at any point (x l9 y x ) 
by the method of the present paragraph, and show that your 
result agrees with that of § 1, Example 2. 

Find the equation of the tangent to each of the following 
curves at the point designated, applying each time the method 
of the text, Reduce the equation to its simplest form. 

12. The curve of Ex. 1 at the point (a^, y x ). 

Ans. 2 x L x + 3 y x y = 12. 

13. The curve of Ex. 3 at the points (x 1} y{) ; (3, — 6). 

14. The curve of Ex. o at the point (x L , y x ). 

Ans. Ax x x + Byiy = C. 

15. The curve of Ex. 6 at the point (x u y^). 

16. The curve of Ex. 7 at the points (x 1} y{) ; (-J-, 1). 

17. The curve of Ex. 9 at the origin. 

18. Find the equations of the normals to the curves of Exs. 
12, 13 at the points specified. 

3. The Equation u + kv = Q. Consider the following ex- 
ample. 

The equations 

(1) * + y-2 = Q, 

(2) x-y = 0, 
represent two straight lines intersecting 
in the point (1, 1), as shown in Fig. 3. 
What can we say concerning the curve * 

(3) (x + y-2)+k(x-y)=0, 
where 7c denotes a constant number ? 

This curve is a straight line, since (3) is an equation of the 
first degree in x and y. Suppose, now, that various different 
values are given to fc. Then (3) represents various straight 
lines in turn. What do all these lines have in common ? 

* The word "curve" is used here in the sense common in analytic 
geometry, to denote merely the " locus of the equation.' 1 Consequently 
a curve in this sense is not necessarily crooked ; it may be a straight line. 




Fig. 3 



1(36 ANALYTIC GEOMETRY 

Since the lines (1) and (2) intersect in the point (1, 1), the 
coordinates of this point make the left-hand sides of equations 
(1) and (2), namely, the expressions, 

x -f y — 2 and x — y, 

vanish. Consequently, they always make the left-hand side 
of equation (3) vanish. In other words, equation (3) is satisfied 
by the coordinates of the point of intersection of the lines (1) and 
(2), no matter what value k has. This means that all the 
straight lines represented by (3) go through the point of inter- 
section of the lines (1) and (2). 

The result can be restated in the following form. Let the 
single letter u stand for the whole expression x -f y — 2 : 

u = x+y-2, 

the sign = meaning identically equal, i.e. equal, no matter what 
values x and y have. Similarly, let v stand for x — y : 

v = x — y. 
Then (3) takes on the form : 
(4) u + kv = 0. 

We now restate our result. 

Ifu = and v = are the equations of tivo intersecting straight 
lines, then the equation 

u + kv = Q 

represents a straight line which goes through the point of inter- 
section of the two given lines. 

By giving to k a suitable value, u -f- kv = can be made to 
represent any desired line through the point of intersection 
( x h yd of the given lines, with the sole exception of the line 
v = 0. For, let L be the desired line, and let (x 2 , y 2 ) be a 
point of L distinct from (x 1} y x ). Then, on substituting for x 
and y the values x 2 and y 2 in the equation u + kv = 0, we ob- 
tain an equation, in which k is the unknown. This equation 
can be solved for k, since v does not vanish for the point 
O2, 2/2). 



CERTAIN GENERAL METHODS 167 

Example. Find the equation of the line L which goes 
through the point of intersection of the lines (1) and (2) and 
cuts the axis of y in the point (0, — 4). 

The required line, L, is one of the lines (3) ; i.e. for a suit- 
able value of k, (3) will represent L. To find this value of k, 
we demand that (3) contain the given point (0, —4) of L. 
We have, then, setting x — and y = — 4 in (3) : 

(0-4-2) + &(0 + 4)=0 or fc = f. 

Consequently, the equation of the line L is 

x + y — 2 + f (x — y)= or 5x — y — 4 = 0. 

That the line represented by the latter equation does actually 
go through the points (1, 1) and (0, — 4) can be verified 
directly. 

The principle which has been set forth for two straight lines 
evidently applies to any two intersecting curves whatever, so 
that we are now in a position to state the following general 
theorem. 

Theorem 1. Let u = and v = be the equations of any 
tivo intersecting curves. TJien the equation 

u + kv = 0, k =£ 0, 

represents, in general,* a curve ivhich passes through all the 
points of intersection of the two given curves, and has )io other 
point in common with either of them. 

The last statement in the theorem is new. To prove it, 
we have but to note that, if the coordinates of a point P satisfy 
the equation u + kv = and also, for example, v = 0, they 
must satisfy the equation u = ; that is, if P is a point on the 
curve u + kv = 0, which lies on one of the given curves, it 
lies also on the other and so is a point of intersection of the 
two. 

* It may happen in special cases that the locus u + kv = reduces to 
a point, as when, for example, 

U = 2 X2 + 2 ^ _ x, U = X2 + 2/2 _ x , fe = — 1, 



168 ANALYTIC GEOMETRY 

Suppose, now, that the equations u = and v = represent 
two curves which have no point of intersection. It follows, 
then, from the argument just given, that the curve 

u + kv = k^=0 

has no point in common with either of the given curves. But 
it may happen, in this case, that there are no points at all 
whose coordinates satisfy the equation u + kv = 0. Thus, if 

u = x 2 -f y 2 — 1, 

v = x 2 + y 2 — 4, 
and k = — 1, we have 

u + kv = 3, 

and there are no points whose coordinates satisfy the equation 
3 = 0. 

The general result can be stated as 

Theorem 2. Let u = and v = be the equations of two 
non-intersecting curves. Then the equation 

u-\-kv — 0, k=£0, 

represents, in general, a curve not meeting either of the two given 
curves. In particular, it may happen that the equation has no 
locus.* 

In the special case that u and v are linear expressions in 
x and y, it is possible to say more. 

Ifu = and v = are the equations of two parallel straight 
lines, the equation 

u + kv = 0, k^=0, 

represents, in general, a straight line parallel to the given lines. 
For a single value of k, the equation has no locus. 
Thus, if the parallel lines are 

u = x-\-y = 0, v = x + y + l = 0, 
the equation 
(5) u + kv =(1 + k)x +(1 + k)y + k = 

* It may happen, also, that the equation represents just one point, as 
when, for example, 

u = x 2 + ?y 2 — 2. v = x* + y 2 — 1. k = — 2, 



CERTAIN GENERAL METHODS 169 

lias no locus when k = — 1, but otherwise it represents a line, 
of slope — 1. parallel to the given lines. In fact, it yields all 
the lines of slope — 1, except the line v = 0, since, if we re- 
write it in the form, 

the quantity k/(l -+- k) may be made to take on any value, ex 
cept 1, by suitably choosing k. 

Pencils of Curves. All the lines through a point, or all the 
parallel lines with a given slope, form what is called a pencil 
of lines. Equation (5) represents, when k is considered as an 
arbitrary constant, all the lines of slope — 1, except the line 

in this case, then, u -f kv = and v = together represent all 
the lines of slope — 1, that is, a pencil of parallel lines. 

Similarly, u + kv = 0, when u = and v = are the lines 
(1) and (2), yields all the lines through the point (1, 1), 
except the line (2) ; hence u + kv = and v — together 
represent all the lines through the point (1, 1), — a pencil of 
intersecting lines. 

Thus, if u = and v = are any two lines, the equations 

(6) u + kv = and v = 

together represent a pencil of lines. 

If we set k = m/l in u -{-kv = and multiply by I, the re- 
sulting equation 

(7) hi + mv = 

is equivalent to the equation a -f kv — when I ^ 0, and when 
1 = (m j= 0), it becomes v = 0. Consequently, the two equa- 
tions (6) may be replaced by the single equation (7). 

The pencil of lines through the point (1, 1), for example, 
may now be given by the single equation 

l(x + y-2)+m(x-y)=0, 

where I and m have arbitrary values, not both zero. 



170 



ANALYTIC GEOMETRY 



In general, if u = and v = are any two curves, all the 
curves represented by the equation 

lu + mv — 0, 

where I and m have arbitrary values, not both zero, form what 
is called a pencil of curves. 

Applications. Example 1. Let 

u = x 2 + y ■ -f ax -f fa/ + c = 0, 
w = a^ + ?/< + a'a; + b'y + c' = 0, 

be the equations of any two circles which cut each other. 
Then the equation 

u — v=(a — a')x+(b - V)y+(c— c')=0 

represents a curve which passes through the two points of 
intersection of the circles. But this equation, being linear, 
represents a straight line, and is, therefore, the equation of 
the common chord of the circles. 

The foregoing proof is open to the criticism that conceivably 
we might have 

a-a' = 0, b-b' = 0, 

and then the equation u — v = would not represent a straight 
line. But in that case the circles would be concentric, and we 

have demanded that they cut each 

other. 

Example 2. We can now proye 
the following theorem : Given three 
circles, each pair of which intersect. 
Then their three common chords 
pass through a point, or are parallel. 

Let two of the three given circles 
be those of Example 1, and let the 
equation of the third circle be 



Wy=0 




iv = a 2 + if + a"x + b"y -f c" = 0. 



CERTAIN GENERAL METHODS 171 

Then the equations of the three common chords can be written 
in the form : 

u — v = 0, v — w = 0, w — u — 0. 

Let 

U x = V — W, Vx = W — U, Wx = U — V. 

We observe that the equation, 

(8) Ux 4- Vx 4- Wx = or — w x = u x + v 1} 

holds identically for all values of x and y. Consequently, the 
line Wx = is the same line as 

u x 4 Vi = 0, 

and therefore it passes through the point of intersection of 
ux = and v x = 0, or, if these lines are parallel, is parallel to 
them. Hence the theorem is proved. 

The above proof is a striking example of a powerful method 
of Modern Geometry known as the Method of Abridged Nota- 
tion* By means of this method many theorems, the proofs of 
which would otherwise be intricate, or for whose proof no 
method of attack is readily discerned, can be established with 
great ease. 

EXERCISES 

1. Find the equation of the straight line which passes 
through the origin and the point of intersection of the lines 

2z-32/-2 = 0, 5oj + 2 y + 1 = 0. 

Arts. 12x + y = 0. 

2. Find the equation of the straight line which passes 
through the point (—1, 2) and meets the lines 

x + y = 0, # + 2/4-3 = 0. 
at their point of intersection. 

* The first general development of this method was given by the 
geometer, Julius Plucker, in his Analytisch-geometrische Entwicklungen 
of 1828 and 1831. 



172 ANALYTIC GEOMETRY 

3. Find the equation of the straight line which passes 
through the point of intersection of the lines - 

5x-2y-3 = 0, 4a; + ly - 11 = 

and is parallel to the axis of y. 

4. Find the equation of the straight line which passes 
through the point of intersection of the lines given in Ex. 3 
and makes an angle of 45° with the axis of x. 

5. Find the equation of the straight line which passes 
through the point of intersection of the lines of Ex. 1 and is 
perpendicular to the first of the lines given in Ex. 3. 

Ans. 3$x + 95y + 58 = 0. 

6. The same, if the line is to be parallel instead of per- 
pendicular. 

7. Find the equation of the common chord of the parabolas 

f- -2y + x = 0, 2/2 + 2x -y = 0. 

Ans. x + y = Q. 

8. The same for the parabolas t 

2x"-5x + 2y = 3, 
3a;- + 7a; — 9?/ = 4. 

9. Write the equation of the pencil of curves determined 
by the two curves (a) of Ex. 1; (b) of Ex. 3; (c) of Ex. 7. 

10. What is the equation of the pencil of circles determined 
by the two circles 

x <l + y 2 -2x-l = 0, 

& + y 2 + 4a - 1 = ? 
Draw a figure showing the pencil. Find the equation of that 
circle of the pencil which goes through the point (2, 4). 

11. Find the equation of the pencil of parallel lines (a) of 
slope 1 : (6) of slope — 3 ; (c) of slope Aq. 

Ans. (a) y = x+k. 

12. Find the equation of the pencil of lines through (a) 
the point (0, 0) ; (6) the point (3, 2) ; (c) the point (0, b) ; 
(d) the point (xq, y ). Ans. (a) lx + my = 0. 



CERTAIN GENERAL METHODS 173 

4. The Equation uv = 0. Consider, for example, the equa- 
tion 

(1) x 2 -y 2 = 0. 
Since x 2 — y 2 == (x - y) (x + y), 
it is clear that equation (1) will be satisfied 

(a) if (x, y) lies on the line 

(2) *-2/ = 0; 

(6) if (x, y) lies on the line 

(3) z + 2/ = 0; 

and in no other case. Equation (1), therefore, is equivalent to 
the two equations (2) and (3) taken together, and it represents, 
therefore, the two right lines (2) and (3). 

It is clear from this example that we can generalize and 
say: 

Theorem. The equation 

uv = 
represents those points (x, y) ichich lie on each of the two curves, 

u = 0, v = 0, 

and no others. 

It follows as an immediate consequence of the theorem that 
the equation uvw ... = Q> 

whose left-hand member is the product of any number of 
factors, represents the totality of curves corresponding to the 
individual factors, when these are successively set equal to zero. 

Example. Consider the equation, 

x*-y* = 0. 
Here,* 

*• - y* =(x* - y 2 ) (af- + y 2 ) = (x -y)(z + y) (x 2 + y 2 ). 

* It is true that the following equation is an identity, and so the sign 
= instead of = might be expected. The use of the sign = for an identical 
equation is not, however, considered obligatory, the sign = being used 
when it is clear that the equation is an identity, so that the fact does 
not require special emphasis. 



174 ANALYTIC GEOMETRY 

The given equation is, therefore, equivalent to the three equa- 

x-y = 0, x + y = 0, a? + ^ = P. 

The first two of these equations represent right lines. The 
third is satisfied by the coordinates of a single point, the 
origin. Since this point lies on the right lines, the third 
equation contributes nothing new to the locus. 

EXERCISES 

What are the loci of the following equations ? 

3. 2x i + 3xy-2y 2 = 0. 4. xy + x + 2y + 2 = 0. 

5. x" 1 -f xy — '2x — 2y = 0. 6. x 3 -f xy 2 = x. 
7. 3x*y-2xy = 0. 8. x A - y* - 2x 2 + 2y* = Q. 

9. (x + y — 1) (x 2 + $^)= 0. ^r*s. The line whose inter- 
cepts on the axes are both 1, and the origin. 

io. (*+aflj(«p+y + i)=<ji 

11. (» + y)t(a-l)* + y»]=a 

12. a; 3 + x n y — &?/ 2 — ?/ 3 = 0. 

Find, in each of the following exercises, a single equation 
whose locus is the same as that of the given systems of equa- 
tions. 

13. a;-2 = 0, y — 4 = 0. 

14. x = 2, y = 4. 

15. aj + y — 2 = 0, x-y + 2 = <\ 

16. x—3y=5, 4<c + 3 = 0. 

.„ x y x y 

17. - = |, - = -f- 
a o a b 

5. Tangents with a Given Slope. Discriminant of a Quad 
ratic Equation. From elementary algebra we know that the 
roots of the quadratic equation 



CERTAIN GENERAL METHODS 



175 



(1) 
are 



Ax 2 + Bx + = 0, 



1 2A^2A 



A^O, 



X = 



2A 2A 



JL-VB 2 -4;Aa 



From these formulas the truth of the following theorem at 
once becomes apparent. 

Theorem 1. The roots of the quadratic equation (1) are 
equal if and only if 

B 2 -±AC=0. 

The quantity B 2 — 4 AC is known as the discriminant of the 
quadratic equation (1). 

By means of the theorem we shall solve the following prob- 
lem. 

Problem. Let it be required to find the equation of the 
tangent to the parabola 

(2) y = 6x, 

which is of slope -|-. 

Let I be a line of slope £-. 
which meets the parabola in 
two points, P 1 and P 2 . If we 
allow L to move parallel to 
itself toward the tangent, T, 
the points P x and P 2 will move 
along the curve toward P, the 
point of tangency of T; and 

if L approach T as its limit, the points P x and P 2 will approach 
the one point P as their limit. 

It is clear that these considerations are valid for any conic. 
Accordingly, we may state the following theorem. 

Theorem 2. A line which meets a conic intersects it in 
general in two points. If these two points approach coincidence 




Fig. 5 



176 ANALYTIC GEOMETRY 

in a single point, the limiting position of the line is a tangent to 
the conic* 

In applying Theorem 2 to the problem in hand, let us denote 
the intercept of the tangent T on the axis of y by ft. The 
equation of T is, then, 

(3) y = -lx + ft. 

The coordinates of the point P, in which T is tangent to the 
parabola, are obtained by solving equations (2) and (3) simul- 
taneously. Substituting in (2) the value of y given by (3), 
we have 

or 

(4) aj2 + 4G8-6)aj + 4£2 = 0. 

The roots of equation (4) are equal, since they are both the 
abscissa of P. Accordingly, by Theorem 1, the discriminant 
of (4) is zero. Hence 

16 (ft - 6) 2 - 16^ = 0, or - 120 + 36 = 0. 

Thus ft = 3, and the tangent to the parabola (2) whose slope 
is \ has the equation 

(5) a?-2y + 6 = 0. 

If in (4) we set ft = 3, the resulting equation, 
x i- 12 x +36 = 0, 

has equal roots, as it should. The common value is x — 6, and 
the corresponding value of y, from (2), is y = 6. The coordi- 
nates of the point of tangency, P, are, then, (6, 6). 

Second Method. We proceed now to give a second method 
of solution for the type of problem just discussed. Let the 
conic be the ellipse 

(6) 4tf + y» = 6, 
and let the given slope be 4. 

* A tangent to a conic might then be defined as the limiting position 
of a line having two points of intersection with the conic, when these 
points approach coincidence in a single point ; this is a generalization of 



CERTAIN GENERAL METHODS 



177 




It is evident from the figure that there are two tangents of 
slope 4 to the ellipse. Let the intercept on the axis of y of 
one of the tangents be ft. The equation of 
this tangent is then 

(7) y = ±x+p. 
Our problem now is to determine the value 

of ft. To this end, let the coordinates of the 
point of contact of the tangent be (a^, y ± ). 
Then a second equation of the tangent is, 
by (12), §2, 

(8) ±x x x + y x y = 5. 
Since equations (7) and (8), which we 

rewrite as 

±x-y + ft = 0, 
4tx& + y$ — 5 = 0, 
represent the same line, it follows, from Ch. II, § 10, Th. 5, 
that 

4ag = 2/i _ -5 
4- -1 ft 

From the equality of the first and third ratios we have 

5 

Since the second and third ratios are equal, 

5 



Fig. 6 



(9) 



z,= 



(10) 



2/i = 



P 



Furthermore, the point (xi, y{) lies on the ellipse and so the 
values of x 1 and y x , given by (9) and (10), satisfy equation (6). 
Accordingly, 

100 , 25 K 25 - 
=5, or — = 1. 

Hence ft has the value 5 or — 5. 

the definition of § 1. A tangent cannot be defined as a line meeting the 
conic in a single point, for there are lines of this character which are not 
tangents, viz., a line parallel to the axis of a parabola, or to an asymptote 
of a hyperbola. 



178 ANALYTIC GEOMETRY 

Substituting these values of fi in turn in (7), we obtain 

4sc — y + 5 = 0, 4# — y — 5 = 0, 

as the equations of the two tangents of slope 4 to the ellipse 
(6). From equations (9) and (10) it follows that the points of 
contact of these tangents are, respectively, (—1, 1) and 

(i- - i). 

Both the methods described in this paragraph are general 
in application. For the usual type of problem met with in a 
first course in Analytic Geometry either method may be used 
with facility. It is, however, to be noted that the second 
method presupposes that the equation of the tangent to the 
curve at an arbitrary point on the curve is known, whereas 
the first does not. Accordingly, in case a curve is given, for 
which the general equation of the tangent is not known, — for 
example, the parabola, y =3 a? 2 — 2a? + 1, — the first method 
will be shorter to apply. 

EXERCISES 

Determine in each of the following cases how many tan- 
gents there are to the given conic with the given slope. Find 
the equations of the tangents and the coordinates of the points 
of tangency. Use both methods in Exs. 1, 2, 3, checking the 
results of one by those of the other. 





Conic 






Slope 


1. 


afi + tf = 5, 






2. 




jLns. 


\2x- 
\2x- 


-y 


— 5 = 0, tangent at (2, — 1), 






-y 


-{-5 = 0, tangent at (—2, 1). 


2. 


tf = 3x, 






f 


3. 


2^ + ^ = 11, 






-i 


4. 


aj 2 + 82/ = 0, 






2. 


5. 


4^-2/ 2 = 20, 






3. 


6. 


%'* + f + 2x = 0, 






i 


7. 


6y?-5x = 0, 






1^ 

X 3' 



CERTAIN GENERAL METHODS 179 

8. What are the equations of the tangents to the circle 

which are parallel to the line Sx — ?/ + 5 = ? 

9. What is the equation of the tangent to the ellipse 

Ax 2 + oy 2 = 20, 

which is perpendicular to the line x + 3y — 3 = and has a 
positive intercept on the axis of y ? 

10. Find the equation of the tangent to the parabola 

y = 3x^ — 2x + 1, 

which is perpendicular to the line x -f- Ay + 3 = 0. 

Arts. 4 x — y — * = 0. 

11. Make clear geometrically that, no matter what direction 
is chosen, there are always two tangents to a given ellipse, 
which have that direction. 

12. How many tangents are there to the parabola y — 2 rase, 
which have the slope 0? State a general theorem relating 
to the number of tangents to a parabola which have a given 
slope. 

13. Are there any tangents of slope 3 to the hyperbola 

4z 2 -y 2 = 5? 
If so, what are their equations ? 

14. The preceding exercise, if the given slope is (a) 1; 
(6) 2. Give reasons for your answers. 

6. General Formulas for Tangents with a Given Slope. 

Consider first the hyperbola 

(i) t-yl =1 . 

Before attempting to find a general formula for the equations 
of the tangents to the hyperbola, which have a given slope, A, 
we shall do well to ask if such tangents exist. In answer to 
this question we state the following theorem. 



180 



ANALYTIC GEOMETRY 



Theorem. All the tangents to the hyperbola (1) are steeper 
than the asymptotes. Their slopes A all satisfy the inequality 



(2) 



X|> 



or 



A 2 >- 



Conversely, if A satisfies (2), there are two tangents of slope A 
to (1). If 9 however, \ 2 <b 2 /a?, there are no tangents of slope A 

«o(i). 

To prove the theorem, let a point P, starting from the vertex 
A, trace the upper half of the right-hand branch of (1). Then 

the tangent, T, at P, starting from 
the vertical position at A, turns 
continuously in one direction, and, 
as Precedes indefinitely, approaches 
the asymptote S as its limit. In 
other words, the slope, A, of T 
decreases continuously through all 
positive values greater than the 
slope, b/a, of S, and approaches b/a 
as its limit.* Consequently, A is always greater than b/a: 




Fig. 7 



a 

* The geometrical evidence of this is convincing, but not conclusive. 
To clinch it, we give the following analytical proof : If the coordinates 
of P are (x, y), the slope \ of T is, by (11), § 2, 



According to Ch. VIH, § 4, eq. (3), 



V b 



Vt-S 



Hence 



x=*. J 



°a/i-- 



When P traces the upper half of the right-hand branch of (1) and re- 
cedes indefinitely, x increases continuously from the value a through all 
values greater than a. Then a 2 /x 2 decreases continuously from 1 and 
approaches as its limit ; and 1 — a 2 /x 2 , and hence v'l — a 2 /x 2 , in- 



CERTAIN GENERAL METHODS 181 

If P now traces the lower half of the right-hand branch, A 
is negative, and always : 

-x> b - 

a 

These two inequalities can be combined into the single in- 
equality (2). Thus (2) is satisfied by the slope A of every tan- 
gent to the right-hand branch of (1), and hence also, because 
of the symmetry of the curve, by the slope A of every tangent 
to the left-hand branch. 

From the reasoning given in the first case, when P traces 
the upper half of the right-hand branch of (1), it follows, not 
only that A > b/a, but also that A takes on every value greater 
than b/a. Hence, if a value of A, greater than b/a, is arbi- 
trarily chosen, there is surely at least one tangent of this 
slope A to (1), and consequently, because of the symmetry of 
the curve, there are actually two. Similarly, if a value of A 
less than — b/a is given. 

To find the equations of the two tangents of slope A to (1), 
in the case that A does satisfy (2), we apply the first of the 
two methods of § 5. Let the equation of one of the tan- 
gents be 

(3) y = \x + /3, 

where (3 is to be determined. Proceeding to solve (1) and (3) 
simultaneously, we substitute for y in (1) its value as given by 

(3) and obtain the equation, 

or 

(4) (&« _ rfxiygfi _ 2 a:p\x - a\W + &) = 0. 

The roots of equation (4) are both equal to the abscissa of the 

creases continuously from and approaches 1 as its limit. Conse- 
quently, the reciprocal, 1/Vl — a 2 /x 2 , of VI — a 2 /x 2 decreases continu- 
ously through all positive values greater than 1 and approaches 1 as its 
limit. Hence, finally, \ decreases continuously through all positive values 
greater than b/a and approaches b/a as its limit, q. e. d. 



182 ANALYTIC GEOMETRY 

point of contact of the tangent (3), and hence the discriminant 
of (4) must vanish. We have, then, 

4 a*/3W + 4a ■ (&'- + /? ) (b ■ - cW) = 0, 

or, simplifying, 

(5) £2 = a 2 * 2 - 6 2 . 

Hence £ has either of the values 



± Va 2 A 2 — 5 2 , 
and the equations of the two tangents, written together, are 

(6) • y = \x ± Va 'A 2 — b \ 

Since A satisfies (2), or the equivalent inequality cPX? — V- > 0, 
the quantity under the radical is positive and so has a square 
root.* We have thus obtained the following result. 

Tlie equations of the tangents to the hyperbola (1), which have 
the given slope A, where A satisfies the inequality (2), are given 
by (6). 

Let the student deduce the following results, using either of 
the two methods of § 5. 

The equations of the tangents to the ellipse 

(7) £ + £=1, 

v } a? ft 2 

which have an arbitrarily given slope A, are 

(8) y = \x± Va ! A? + bl 
The equation of the tangent to the parabola 

(9) y 2 = 2mx, 
which has a given slope A, not 0, is 

(10) y = *x + ~ 



*If we take a value of X, for which X 2 < 6 2 /a 2 , " then a 2 X 2 - 6 2 is 
negative and has no square root. Consequently, there are no tangents 
with this slope, as the theorem states. Finally, if \ = ±b/a, then 
a 2\2 _ f)i — o 5 and (4) is not a quadratic equation. 



CERTAIN GENERAL METHODS 183 

Condition that a Line be Tangent to a Conic. The two 
methods used to find the tangent to a conic with a given slope 
apply equally well to the problem of determining the condi- 
tion that an arbitrary line be tangent to a given conic. In 
fact, in finding the equations of the tangents of slope X to the 
hyperbola (1), we have at the same time shown that the con- 
dition that the line 

(11) y = Xx + ft . 

where u:e now consider X and /? both arbitrary, be tangent to the 
hyperbola (1), is that X and /? satisfy the equation (5) : 

Similarly, the work of deriving formula (8) or (10) involves 
finding the condition that the line (11) be tangent to the ellipse 
(7) or the parabola (9). 

Example. Is the line 3x — 2 ?/ -h 5 = tangent to the hyper- 
bola a- -4^ = 4? 

It is, if, when we write the equations of the line and the 
hyperbola in the forms (11) and (1), the values which we 
obtain for X, ft a\ and 5 ? , namely, f, f, 4, and 1, satisfy (5). 
It is seen that they do not, and hence the line is not tangent 
to the hyperbola. 

EXERCISES 

1. Derive formula (8) and at the same time show that the 
condition that the line (11), where now A. and ft are both arbi- 
trary, be tangent to the ellipse (7) is that X and /3 satisfy the 
equation 

(12) ft = a 2 A 2 + V. 

2. Show that the line (11) is tangent to the parabola (9) if 
and only if 

(13) 2 X/3 = m. 
Hence prove the validity of formula (10). 



184 ANALYTIC GEOMETRY 

3. By direct application of the methods of the text, show 
that the condition that the line (11) be tangent to the circle 

(14) xi+tf^a? 
is that 

(15) /^ = a 2 (l + A*). 

4. Using formulas (6), (8), and (10), find the equations of 
the tangents which are required in Exs. 1, 2, 3, 5, and 7 of § 5. 

5. Has the hyperbola 9 a*-— 4?/ 2 = 36 any tangents whose 
inclination to the axis of x is 60° ? Whose inclination is 45° ? 
If so, find their equations. 

6. Eind the equations of the tangents to the parabola 
y 2 = 8x, one of which is parallel to and the other perpendicu- 
lar to the line 3&* — 2^ + 5 = 0. Show that these tangents 
intersect on the directrix. 

7. Prove that any two perpendicular tangents to a parabola 
intersect on the directrix. 

In each of the following exercises determine whether the 
given line is tangent to the given conic. If it is, find the 
coordinates of the point of contact. 

Conic Line 

8. 2x"< + 3y' l = 5, 2x-Sy-5 = 0. 
9.if = 2x, a*-f-4?/ + 8 = 0. 

10. 3a*<-5^ = 7, 6a -5?/ -8 = 0. 

In each of the following cases the equation of the given line 
contains an arbitrary constant. Eind the value or values of 
this constant, if any exist, for which the line is tangent to the 
given conic. 

Conic Line 

11. a* 2 + 3 ^ = 4, x — 3?/ + c = 0. Ans. c = ±4. 

12. x*- ?/- = 3, 2a* + <fy-3 = 0. 

13. 5?/ = 3a*, 7cx-10y + 15 = 0. 

14. 4 x* — 3 y 2 = 1, x -f 2 y -f k = 0. 



CERTAIN GENERAL METHODS 



185 



P:(-l,2) 



15. Is the line x + y = 1 tangent to the parabola y = x — x 2 ? 

16. Show that the lines 3# ± y + 10 = are common tan- 
gents of the circle a; 1 -f- y 1 = 10 and the parabola y = 120 a;. 

17. Find the equations of the common tangents of the 
parabola y 2 = 4V2# and the ellipse x 1 + 2y 2 = 4. 

Ans. V2a;± 22/ + 4 = 0. 

7. Tangents to a Conic from an External Point. Given a 
point P external to a conic, that is, lying on the convex side of 
the curve. From P it is possible, in general, to draw two 
tangents to the conic. It is required 
to find the equations of these tan- 
gents. 

Let the conic be the ellipse 

(1) tf + 2^ = 3 

and let P be the point (— 1, 2). We 
find the equations of the two tangents 
drawn from P to the ellipse by find- 
ing first the coordinates of the 
points of tangency. Let P x be the 

point of tangency of one of the tangents, and let the coordi- 
nates of P l5 which are as yet unknown, be (a^, y^). The 
equation of this tangent is then, by (12), § 2, 

(2) a*r + 2 W = 3. 

There are two conditions on the point P 1? to serve as a means 
of determining the values of x x and y x . In the first place, the 
tangent (2) at P x must go through the point (— 1, 2) ; hence 

(3) -^ + 4^ = 3. 

Secondly, the point P x lies on the ellipse (1) ; that is, 

(4) ^ + 2^ = 3. 

Equations (3) and (4) are two simultaneous equations in the 
unknowns a^, y v If we solve (3) for x 1 : 

(5) ^ = 4^-3, 




Fig. 8 



186 ANALYTIC GEOMETRY 

and substitute its value in (4), we obtain, on simplification, the 
following equation for y 1 : 

(6) 32^-4^ + 1 = 0. 

The roots of this equation are y x = 1 and y x = \ ; the corre- 
sponding values of x x are, from (5), 1 and — J. Hence (x h y x ) 
= (1, 1) and (x 1} yi) = (— -J, i) are the solutions of (3) and (4). 
The coordinates of the points of tangency are, therefore, 
(1, 1) and (— -§-, £). Substituting the coordinates of each point 
in turn for x x y x in (2) and simplifying the results, we obtain, 
as the equations of the two tangents, 

(7) a + 22/-3 = and 5x- 2y + 9 = 0. 

The method used in this example is universal in its applica- 
tion, not only to conies, but to other curves as well. It should 
be noted, however, that the equation corresponding to (6) does 
not, in general, have rational, that is, fractional or integral, 
roots. Usually its roots involve radicals and hence so do the 
final equations of the tangents. If one were dealing with an 
arbitrary point P external to an arbitrary conic, for example, 
the ellipse 

these radicals would be complicated. Accordingly, we make no 
attempt to set up general formulas for the tangents to a given 
conic from an external point. We have expounded a method 
which is applicable in all cases, and this is the purpose we set 
out to achieve. 

Second Method. We give briefly an alternative method of 
finding the equations of the tangents from the point (— 1, 2) to 
the ellipse (1). 

Suppose one of the tangents is the line 

(8) -y = Aa> + j8. 

Since it is a tangent to (1), we have, according to § 6, Ex. 1, 
20 , = 6A. 2 +-3. 



CERTAIN GENERAL METHODS 187 

Since it contains the point (—1, 2), 

2 = -A+ i 8. 

If we solve these equations in A and /? simultaneously, we find 
that A. = — i or -| and that ft = -| or f. Substituting these pairs 
of values for A and (3 in turn in (8) and simplifying the results, 
we obtain the equations (7). 

EXERCISES 

1. Make clear geometrically that from a point external to 
an ellipse or a parabola there can always be drawn just two 
tangents to the curve. 

2. How many tangents can be drawn to a hyperbola from 
its center ? From a point on an asymptote, not the center ? 
From any other external point ? Summarize your answers in 
the form of a theorem. 

3. Let P be a point external to a hyperbola from which two 
tangents can be drawn to the curve. How must the position of 
P be restricted, if the two tangents are drawn to the same 
branch of the hyperbola? To different branches? 

4. The point (2, 0) is a point internal to the hyperbola 
x l — 2# 2 = 2. Prove analytically that no tangent can be 
drawn from it to the curve. 

In each of the following exercises determine how many tan- 
gents there are from the point to the conic, and when there are 
tangents, find their equations. Use the first method. 



a*. {Z +29 - B =\ 

{ 2x-y-5 = 0. 





Conic 


Point 


5. 


x* + tf = 5, 


(3,1). . 


6. 


a 2 — 3^ = 4, 


(b ~ 1). 


7. 


a,* -2^ = 2, 


(1,-2). 


8. 


4^-9?/ 2 = 36, 


(4, 1). 


9. 


^2 — 4aj = 0, 


(4, 5). 


10. 


a;--42/ 2 = 4, 


(2, 1). 



188 ANALYTIC GEOMETRY 

11. x 2 -8y = 0, (3,2). 

12. 2^ -3^ = -10, (~2,1). 

13. x 2 + y 2 -4:X-y = 0, (5,2). 

14. ^ + 2/ 2 = 25, (-1,7). 

15. Work Exercises 5-10 by the second method. 

16. Show, by use of the second method, that the tangents 
from the point (2, 3) to the ellipse 4 a; 2 + 9y 2 = 36 are perpen- 
dicular. 

EXERCISES ON CHAPTER IX 

1. Prove that the slope of the conic 

(1 - e 2 )x 2 + y 2 -2mx + m 2 = 
at the point (x lf yi) is 

X — _ (I ~ e2 ) x i — m , 

Hence show that the equation of the tangent at (x 1} y x ) is 
(1 — e°) x& + y$ — m (x + a^) + m 2 = 0. 

2. Show that the slope of the curve 

Ax 2 + Bxy + Cy 2 + Dx + Ey + F= 

at the point (x ly y ± ) is 

A= 2^ + ^ + D 

Bx l + 2Gy l -^E' 

Then prove that the equation of the tangent at (x h ?/i) is 
-A*i* + f (2/i* + *#) + Q/i2/ + f (* + *i) + |(2/ + 2/i) + F= 0. 

3. The following equations contain arbitrary constants. 
What does each represent ? 

(a) 2/ = Aa + 3; 

Arts. All the lines through (0, 3) except x = 0. 
(6) 3aa + 22/ + a-3:=0; 

(c) 7a5 + 52/ — c + 3 = 0; 

(d) (2a + 5)a + (7a-3)?/ = 9a + 2; 



CERTAIN GENERAL METHODS 189 

(e) te+(2/ + ra)?/-3ra = 0; 
(/) (22 + 3 m)x -(4Z + 6m)y + 5? = 0. 

4. A line moves so that the sum of the reciprocals of its 
intercepts is constant. Show that it always passes through a 
fixed point. 

5 A line with positive intercepts moves so that the excess 
of the intercept on the axis of x over the intercept on the axis 
of y is equal to the area of the triangle which the line forms 
with the axes. Show that it always passes through a fixed 
point. 

6. Prove that the straight lines, 

5x-2y + 6 = 0> 
2a;-42/ + 3 = 0, 
3x + 2y + 3 = 0, 

meet in a point, by showing that the equation of one of them 
can be written in the form lu + mv = 0, where u = and v = 
are the equations of the others. 

7. Show that the three lines, 

x + 3y- 4 = 0, 
5x-3y+ 6 = 0, 
3a — 9?/ + 14 = 0, 
meet in a point. 

8. Prove that the three lines 

ka — 1(3 = 0, lj$ — my = 0, my — ka = 0, 

where ce=0, /?=0, and y=0 are themselves equations of 
straight lines and k, I, and m are constants, meet in a poiut. 

9. Find the equation of the common chord of the two in- 
tersecting circles 

x*+ ^ + 6a?-8y+ 3 = 0, 
2x> + 2tf- - 3x + Ay - 12 = 0. 

10. Show that the two circles, 

x 2 + 2/ 2_4 a ._4 i/ _io = 0, 
x 2 + y 1 + 6z + 6y + 10 = 0, 



190 ANALYTIC GEOMETRY 

are tangent to one another. Find the equation of the common 
tangent and the coordinates of the common point. 

11. Find the equation of the circle which goes through the 
points of intersection of the two circles of Ex. 9 and through 
the origin. 

12. Find the equation of the circle which is tangent to the 
circles of Ex. 10 at their common point and meets the axis of 
x in the point x = 2. 

13. What is the equation of the circle which passes through 
the points of intersection of the line 

2a,-- # + 4 = 
and the circle 

x 2 + y 2 + 2x — 4y + 1 = 0, 

and goes through the point (1,1)? 

14. Determine the equation of the ellipse which passes 
through the points of intersection of the ellipse 

x 2 + 4 y 2 = 4 
and the line Sx — 4y — 3 = 0, 

and goes through the point (2, 1). By a transformation to 
parallel axes (cf. Ch. XI, § 1), prove that this ellipse has axes 
parallel to those of the given ellipse and has the same 
eccentricity. 

15. Find a single equation representing both diagonals of 
the rectangle whose center is at the origin and one of whose 
vertices is at the point (a, b). 

16. What is the condition that the equation 

aw - by = o 

represent two perpendicular lines ? 

17. Find the locus of each of the following equations : 

(a) 6x 2 + 5xy- 4# 2 = 
(6) 4^-20^ + 25^ = 
(c) x ' + xy+ 6y=0 



CERTAIN GENERAL METHODS 191 

18. Prove that the equation 

(1) Aa?- + Bxy+CY=*Q 

represents the origin, a single straight line, or two straight 
lines, according as the discriminant, B" — 4:AC, is negative, 
zero, or positive. 

19. Show that, if equation (1), Ex. 18, represents two 
straight lines, the slopes of these lines are the roots of the 
equation 

CA 2 + BX + A = 0. 

20. Prove that the equation 

14 a;- ■— 45a;?/ — 14?/ 1 = 
represents two perpendicular straight lines. 

21. Show that equation (1), Ex. 18, represents two perpen- 
dicular straight lines if and only if A 4- C = 0. 

22. Prove that the equation 

•f — 2xysecfr+x = 

represents two straight lines which form with one another the 
angle 6. 

23. A regular hexagon has its center at the origin and two 
vertices on the axis of x. Find a single equation which repre- 
sents all three diagonals. Ans. y 3 — 3 x"y = 0. 

24. Determine the points of contact of the tangents drawn 
to an ellipse from the points on the conjugate axis which are 
at a distance from the center equal to the semi-axis major. 

25. Find the equations of the common tangents of each of 
the following pairs of conies : 

(a) ^ + */ 2 =16, ^ = 6a>; 

(b) ^l + ^l-l ^ + 11-1. 
W 25 + 9 ' 16 + 25 ' 

W 16 9 ' 25 16 

Draw a good figure in each case, showing the common tangents. 



192 ANALYTIC GEOMETRY 

26. Show that the line 

A ' B 



( 2 ) ~ A + 



is tangent to the circle 

x 2 + f- = a 2 

if and only if — + — = -• 

J A 1 & a 2 

27. Find the condition that the line (2), Ex. 26, be tangent 
to the ellipse 

£ + £=!■ ^ S . £+*«L 

a 2 6 2 ^4 2 J3 2 

28. What will the condition obtained in Ex. 27 become in 
ae case of the hyperbola 

a 2 V 

29. Prove that the line (2), Ex. 26, is tangent to the 
parabola 2/^ = 2 mx, if and only if 2B" + ^4ra = 0. 

30. Find the condition that the line y = \x + /3 be tangent 
to the conic 

(1 - e 2 )«2 + ^2 _ 2mx + m 2 = 0. 

Ans. (fi 4- mA) 2 - e 2 (/? 2 + m 2 ) = 0. 

31. In an ellipse there is inscribed a rectangle with sides 
parallel to the axes. In this rectangle there is inscribed a 
second ellipse, with axes along the axes of the first. Show 
that a line joining extremities of the major and minor axes of 
the first ellipse is tangent to the second. 



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